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pDUCTION  TO 
ERMODYNAMICS 


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AN  INTRODUCTION  TO 
THERMODYNAMICS 


FOE  ENGINEERING  STUDENTS 


BY 


JOHN  MILLS 

H 

PROFESSOR   OF   PHYSICS   AND    ELECTRICAL    ENGINEERING 
COLORADO  COLLEGE 


GINN  AND  COMPANY 

BOSTON  •  NEW  YORK  •  CHICAGO  •  LONDON 


COPYRIGHT,  1910 
BY  JOHN  MILLS 


ALL  RIGHTS   RESERVED 
710.9 


at  fa  en  gum 


G1NN   AND  COMPANY-  PRO- 
PRIETORS •  BOSTON  •  U.S.A. 


PEEPACE 


This  book  is  the  outcome  of  the  writer's  desire  to  put  into  the 
hands  of  a  class,  studying  introductory  thermodynamics,  in  a  concise 
form  a  statement  of  those  principles  and  concepts  which  are  essential 
to  the  study  of  the  construction  and  operation  of  steam  engines, 
steam  turbines,  and  their  auxiliaries,  and,  to  a  more  limited  extent,  of 
air  compressors  and  gas  engines.  The  book  deals  only  with  the  ideal 
and  limiting  cases,  and  aims  only  at  a  preparation  of  the  student  for 
reading  the  more  advanced  technical  works  or  papers.  As  such  the 
text  makes  no  claim  to  originality  except,  to  some  extent,  in  the  selec- 
tion and  arrangement  of  material  and  in  the  location  of  the  emphasis. 

The  material  selected  is  from  such  sources  as  the  works  and  books 
of  Clausius,  Rankine,  Carnot,  Kelvin,  Stodola,  Pellat,  Planck,  Preston, 
Edser,  Bryan,  Boulvin,  Swinburne,  Peabody,  and  Buckingham.  The 
arrangement  is  intended  to  lead  the  student  from  the  sophomore 
physics  work  into  thermodynamics  as  a  continuation  and  more  detailed 
study  of  a  portion  of  physical  science  already  somewhat  familiar.  The 
emphasis  is  therefore  placed  upon  the  physical  concepts,  and  constant 
reference  is  made  to  the  molecular  kinetic  theory,  which  is  usually 
familiar  to  such  a  student.  Although  an  effort  has  been  made  to 
eliminate  as  far  as  practicable  the  use  and  solution  of  differential 
equations,  it  is  hoped  that  should  the  book  be  used  by  a  student  who 
later  takes  up  the  study  of  pure  thermodynamics,  he  will  have  little 
if  anything  to  unlearn  and  will  find  the  physical  interpretation  of  the 
mathematical  relations  rendered  easier.  The  emphasis  is  also  placed 
upon  the  solution  of  numerical  problems,  many  examples  of  which  are 
given.  These  in  general  are  arranged  to  lead  from  the  simpler  con- 
cepts to  the  ideal  limiting  cases  ^  of  the  applications  to  engineering 
problems. 


P/iS19101 


iv  PREFACE 

In  addition  to  the  general  statement  of  sources  made  above, 
acknowledgments  of  details  are  made  in  footnotes.  It  is  the  writer's 
pleasure  to  acknowledge  the  influence  of  former  study  under  Professor 
C.  W.  Berry  of  the  Massachusetts  Institute  of  Technology.  To  the 
inspiration  of  Professor  R.  A.  Millikan  of  the  University  of  Chicago, 
a  friend  and  former  teacher,  who  has  read  a  portion  of  the  proof  of 
this  text,  a  general  acknowledgment  is  gratefully  made  for  most  that 
is  good  in  the  writer's  point  of  view  and  pedagogical  method  in 

physics. 

JOHN  MILLS 

COLORADO  SPRINGS,  COLORADO 


CONTENTS 

PAGE 

CHAPTER  I.    FUNDAMENTAL  CONCEPTS  AND  LAWS 1 

1.  Work.  2.  Energy.  3.  Molecular  Kinetic  Theory.  4.  Temperature. 
5.  Thermal  Units.  6.  First  Law  of  Thermodynamics.  7.  Specific  Heat. 
8.  Component  Parts  of  Specific  Heat.  9.  Latent  Heat.  10.  Boyle's  Law. 
11.  Joule's  Law.  12.  Perfect  Gases.  13.  Charles's  Law.  14.  Standard  Hydro- 
gen Thermometer.  15.  ^Absolute  Temperature.  16.  General  Equation  for  a 
Perfect  Gas.  17.  Use  of  the  Term  "Specific."  18.  Specific  Heat  of  Gases. 
19.  Graphical  Representation  of  Gas  Transformations.  20.  Work  Calculations 
for  Gas  Transformations.  21.  Adiabatic  Transformation.  22.  Comparison  of 
Adiabatic  and  Isothermal  Transformations.  23.  Second  Law  of  Thermo- 
dynamics. 24.  Carnot  Cycle.  25.  Reversible  Cycle.  26.  Carnot  Theorem. 
27.  Discussion  of  Reversible  Processes.  28.  Discussion  of  the  Second  Law. 
29.  Thermodynamic  Scale  of  Temperature.  30.  Carnot  Cycle  for  a  Perfect 
Gas.  31.  Entropy.  32.  Measurement  of  Entropy.  33.  Temperature-Entropy 
Plot.  34.  Discussion  of  Carnot  Cycle  from  0-$  Plot.  35.  Entropy  Changes  in 
Irreversible  Processes. 

CHAPTER  II.    GASES 38 

36.  Graphical  Representation  of  State  for  a  Perfect  Gas.  37.  Equations  for 
Polytropic  Transformation  for  a  Perfect  Gas.  38.  Fundamental  Heat  Equations 
for  a  Perfect  Gas.  39.  Entropy  Changes  for  a  Perfect  Gas.  40.  Relation  of  the 
Specific  Heats  of  a  Gas.  41.  Changes  in  Intrinsic  Energy  of  a  Gas.  42.  Imper- 
fect Gases.  43.  Re'sume'  of  Equations  of  Chapters  I  and  II. 

PROBLEMS  AND  SOLUTIONS  :  GASES 46 

1.  Use  of  the  Characteristic  Equation.  2.  Carnot  Cycle.  3.  Intrinsic  Energy. 
4.  Constant-Pressure  Transformation.  5.  Exponent  n.  6.  Entropy  Changes. 
7.  Isothermal  Expansion.  8.  Adiabatic  Expansion.  9.  Simultaneous  Equations 
in  p  and  v.  10.  Pressure-Volume  Plot.  11.  Regenerative  Cycle,  Sterling  En- 
gine. 12.  Air  Compressor  with  and  without  Clearance.  13.  Otto  Gas  Engine 
Cycle.  14.  Diesel  Internal  Combustion  Cycle.  15.  Air  Refrigeration. 

CHAPTER  III.    WATER  AND  ITS  SATURATED  VAPOR 61 

44.  Phenomena  of  Vaporization.  45.  Specific  Heat  of  Water.  46.  Heat  of 
Vaporization.  47.  Quality  of  Steam.  48.  Intrinsic  Energy  of  Steam.  49.  En- 
tropy of  the  Liquid.  50.  Entropy  of  Vaporization.  51.  Temperature-Entropy 
Chart  for  Steam.  52.  Isentropic  Transformation  of  a  Mixture.  53.  Specific 
Heat  of  Steam.  54.  Re'sume'  of  Equations  for  a  Mixture  of  Steam  and  Water. 

PROBLEMS  AND  SOLUTIONS  :  SATURATED  WATER  VAPOR 75 

16.  Quality.  17.  Intrinsic  Energy.  18.  Isoenergic  Transformation.  19.  Iso- 
thermal Transformation.  20.  Adiabatic  Transformation.  21.  Constant-Volume 
Transformation.  22.  Boiler  Explosion.  23.  Equivalent  Evaporation  and  Boiler 


vi  CONTENTS 

PAGE 

Horse  Power.  24.  Surface  Condenser.  25.  Jet  Condenser.  26.  Rankine  Cycle. 
27.  Incomplete  Expansion.  28.  Efficiency  and  Boiler  Pressure.  29.  Condensing 
Engine.  30.  Use  of  the  Temperature-Entropy  Chart.  31.  Boulvin  Diagram. 
32.  Hyperbolic  Expansion. 

CHAPTER  IV.    SUPERHEATED  STEAM 92 

55.  Three  Molecular  States.  56.  Superheated  Steam.  57.  Specific  Heat  of 
Superheated  Steam.  58.  Entropy  of  Superheated  Steam.  59.  Temperature- 
Entropy  Chart  for  Superheated  Steam.  60.  Intrinsic  Energy  of  Superheated 
Steam.  61.  Isentropic  Transformation.  62.  Superheating  in  Engineering  Prac- 
tice. 63.  Re'sume'  of  Equations  for  Superheated  Steam. 

PROBLEMS  AND  SOLUTIONS  :  SUPERHEATED  STEAM 100 

33.  Specific  Volume.  34.  Temperature  of  Superheat.  35.  Specific  Heat  of 
Superheated  Steam.  36.  Total  Heat.  37.  Entropy.  38.  Quality.  39.  Quality 
after  Adiabatic  Expansion.  40.  Intrinsic  Energy.  41.  Work  of  Adiabatic  Ex- 
pansion. 42.  Gain  by  Superheating. 

CHAPTER  V.    FLOW  OF  STEAM  AND  GASES 105 

64.  Formula  of  de  Saint-Venant.  65.  Alternative  Expression  for  the  For- 
mula of  de  Saint-Venant.  66.  Zeuner's  Form-ula  for  Kinetic  Energy.  67.  Ki- 
netic Energy  of  an  Irreversible  Adiabatic  Expansion.  68.  Loss  of  Kinetic 
Energy  due  to  Friction.  69.  Flow  through  an  Orifice.  70".  Fliegner's  Formulas 
for  the  Flow  of  Air.  71.  Re'sume'  of  Equations  for  the  Flow  of  Fluids. 

PROBLEMS  AND  SOLUTIONS  :  FLOW  OF  FLUIDS 124 

43.  Kinetic  Energy  of  Steam.  44.  Kinetic  Energy  of  Superheated  Steam. 
45.  Loss  due  to  Friction.  46.  Flow  of  Air.  47  and  48.  Napier's  Formula. 
49.  Nonlifting  Injector.  50.  Peabody  Throttling  Calorimeter.  51.  Limits  of 
the  Throttling  Calorimeter.  52.  Throttle  Control.  53.  Nozzle  Area.  54.  Tur- 
bine Nozzle. 

MISCELLANEOUS  PROBLEMS 125 

TABLES 129 

1  and  2.  Gas  Constants.     3.  Steam  and  Water  Constants.     4.  Logarithms. 

INDEX  ..  135 


SYMBOLS 


The  symbols  given  below  are  used  consistently  throughout  this  text.  They  have 
been  selected,  as  far  as  is  possible,  in  accordance  with  the  following  rules  :  (1)  a  given 
symbol  is  to  be  used  /for  but  one  physical  magnitude,  (2)  a  lower-case  letter  is  always 
to  represent  the  value  per  unit  mass  of  the  magnitude  to  which  it  refers,  (3)  the  sym- 
bols are  to  be  in  accordance  with  the  American  usage. 


A  =  l/J=  factor  for  reducing  me- 
chanical units  of  energy  to 
heat  units. 

a  =  general  expression  for  area. 

a  =  volume  coefficient  of  a  gas 

=  1.(*Z.\ 

V\dt   ), 

Apu  =  external  latent  heat  of  vaporiza- 
tion in  thermal  units. 
B.t.u.  =  British  thermal  unit. 

/3  =  pressure  coefficient  of  a  gas 


I  = 


p  \dt 
general    expression    for   specific 

heat. 
specific    heat   at  constant   pres- 

sure. 

specific  heat  at  constant  volume. 
total  intrinsic  energy  of  a  sub- 

stance. 
intrinsic  energy  of  a  unit  mass  of 

a  substance. 
thermal  efficiency  of  a  process  or 

cycle. 

general  expression  for  force. 
numerical  value  of  the  accelera- 

tion due  to  gravity. 
general  expression  for  density. 
heat  units  required  to  superheat 

from  dry  saturated  steam,  per 

unit  mass. 
intrinsic  energy  due  to  molecular 

configuration. 
factor  for  reducing  thermal  units 

to  work  units. 
ratio  of  cp  to  cv. 
loss  of  kinetic  energy  due  to  fric- 

tion,  per  unit  mass  of  steam. 


A  =  total   heat   contents  per    unit  mass 

of  a  mixture    (from  32°  F.),   e.g. 

A  =  q  +  r  +  h,    or    \  =  q  +  xr,    or 

A  •  =  q  +  xp  +  xApu. 

M  =  mass  (or  weight)  per  second  in  flow 

of  fluids. 

m  =  mass  (or  weight). 
n  =  general  expression  for  the  exponent 

in  the  relation  pvn  =  constant. 
p  =  general  expression  for  pressure  (i.e. 

specific  pressure). 
$  =  general  expression  for  entropy. 
<f>w  =  general    expression    for  entropy  of 
liquid  (per  unit  mass)  above  that 
of  water  at  32°  F. 
0v  =  increase   in   entropy  per  unit  mass 

due  to  vaporization. 
<t>s  =  increase   in  entropy  per  unit   mass 

due  to  superheating. 
q  =  heat  contents  of  water  per  unit  mass 

measured  from  32°  F. 
£l  —  general    expression  for  heat  trans- 
ferred   to    or    from    an    external 
source. 

q  =  general    expression  for  heat  trans- 
ferred per  unit  mass. 
R  =  constant  in  the  gas  relation 

p  V  =  mRd. 
r  =  total  latent  heat  of  vaporization  per 

unit  mass. 
p  =  internal  latent  heat  of  vaporization 

per  unit  mass. 
S  =  intrinsic    energy   due   to   molecular 

motion. 
s  =  specific    volume    of    dry    saturated 

steam. 

<r  =  specific  volume  of  water =0.016  cubic 
foot  per  pound. 


vii 


Vlll 


THERMODYNAMICS 


T  =  temperature  measured  on  the  ther- 

modynainic  scale. 

t  =  general  expression  for  temperature. 
t  =  general  expression  for  time. 
0  =  absolute  temperature  as  measured  on 

a  perfect  gas  scale. 
t0  =  saturation  temperature  of  dry  steam 

before  superheating. 
00  =  absolute   saturation   temperature  of 

dry  steam  before  superheating. 
V  —  general  expression  for  volume. 
v  =  general  expression  for  the  volume  of 

a  mixture  per  unit  mass. 
ft  =  velocity. 


u  =  increase  in  specific   volume  due  to 

vaporization,  i.e.  u  =  s  —  <r. 
W  —  general  expression  for  external  me- 
chanical work. 

w  —  general  expression  for  external  work 
per  unit  mass  of  working  sub- 
stance! 

x  =  quality  of  a  mixture  of  x  parts  steam 
and  (1  —  cc)  parts  water. 

y  =  friction    work   during  flow  through 

a  channel  per  unit  mass  of  the  fluid. 

z  and  dz  =  distance  through  which  a  force 

acts,  e.g.  W  =  \  Fdz. 


AN  INTRODUCTION  TO 
THERMODYNAMICS 

CHAPTER   I 

FUNDAMENTAL  CONCEPTS   AND  LAWS 

1.  Work.     Work  is  defined  as  the  act  of  overcoming  a  resistance 
through  space.     It  is  measured  by  the  product  of  the  component  of 
the  acting  force  along  the  direction  of  the  motion  of  its  point  of 
application,  and  the  distance  through  which  its  point  of  application 
moves  during  its  action.     The  scholium  to  Newton's  Third  Law  of 
Motion  states  that  "  if  the  action  (activity)  of  an  agent  be  measured 
by  its  amount  and  its  velocity  conjointly,  and  if  similarly  the  reac- 
tion (counter-activity)  of  the  resistance  be  measured  by  the  veloci- 
ties of  its  several  parts  and  their  amounts  conjointly,  whether  these 
arise  from  friction,  molecular  force,  weight,  or  acceleration,  action 
and  reaction  in  all  combinations  of  machines  will  be  equal  and  op- 
posite."    Since  velocity  is  merely  the  rate  at  which  space  is  traversed, 
it  is  seen  that  this  scholium  states  that  during  any  given  time  the 
work  of  the  acting  forces  is  equal  to  the  work  of  the  resisting  forces. 

The  unit  of  work  depends  upon  the  chosen  unit  of  force.  In  the 
absolute  C.  G.  S.  system  where  the  unit  of  force  is  the  dyne  (a  force 
which  if  acting  constantly  upon  one  gram  will  produce  a  change  in 
its  velocity  of  one  centimeter  per  second  every  second),  the  unit  of 
work  is  the  erg,  or  the  work  done  by  one  dyne  acting  through  one 
centimeter.  In  the  gravitational  C.  G.  S.  system  the  unit  of  force  is 
the  gram  weight  (i.e.  980  dynes),  and  the  unit  of  work  is  the  gram- 
centimeter.  The  force  unit  in  the  F.  P.  S.  system  (gravitational)  is 
the  pound,  or  the  force  of  the  earth's  attraction  upon  one  pound  of 
matter,  and  the  unit  of  work  is  the  foot  pound. 

2.  Energy.     Energy  is  defined  as  the  capacity  of  a  body  for  doing 
work.     It  is  evident,  then,  that  work  and  energy  are  to  be  measured 
in  the  same  units.     In  general  the  energy  possessed  by  a  body  or  by 
a  system  of  bodies  may  be  considered  as  the  result  of  work  done 
upon  that  body  or  system  by  another  body  or  system.     The  energy 
of  a  system  may  be  either  its  capacity  to  do  work  as  the  result  of 

1 


2  THEKMODYNAMICS 

work  done  upon  the  system  in  giving  to  its  component  parts  motion 
(that  is,  work  done  against  the  resistance  of  acceleration),  in  which 
case  it  is  known  as  kinetic  energy ;  or,  it  may  be  the  result  of  work 
done  upon  the  system  in  producing  the  existing  configuration  of  its 
component  parts  (that  is,  work  done  against  gravitation  or  other 
forces),  in  which  case  it  is  known  as  potential  energy. 

The  sum  of  the  potential  and  kinetic  energy  of  a  body  is  com- 
monly called  its  intrinsic  energy.  The  Principle  of  the  Conserva- 
tion of  Energy  states  that  if  work  is  done  upon  a  body  the  amount 
of  work  is  equal  to  the  increase  in  intrinsic  energy  of  the  body ;  that 
is,  to  the  sum  of  the  increases  in  the  kinetic  and  potential  energy. 
And  conversely,  the  work  done  by  a  body  is  equal  to  the  decrease  in 
its  intrinsic  energy.  Thus,  if  dE  represents  the  increase  in  the 
intrinsic  energy  of  a  body  and  dWthe  amount  of  work  done  by  the 
body,  then  dE  +  d  W  =  0. 

The  measure  of  the  kinetic  energy  of  a  rigid  body  is  the  sum  of 
its  kinetic  energy  of  translation  and  its  kinetic  energy  of  rotation. 
The  kinetic  energy  of  translation  is  ^  wt)02,  where  m  is  the  mass  of 
the  body  and  t)0  is  its  absolute  velocity  of  translation.  The  kinetic 
energy  of  rotation  is  ^Ka>^,  where  ^Tis  the  moment  of  inertia  of  the 
body  about  an  axis  through  its  center  of  mass,  and  o>0  is  the  absolute 
velocity  of  rotation  about  this  axis.  The  potential  energy  of  a  body 

is   1  Fdz,  where  the  integral  represents  the  work  done  against  the 

resisting  forces  in  moving  the  body  from  a  position  of  zero  potential 
energy  to  the  position  under  consideration. 

From  the  above  statement  of  the  measure  of  the  intrinsic  energy 
of  a  body  it  is  evident  that  its  energy  is  the  total  amount  of  work 
which  could  be  derived  from  it  under  the  most  favorable  theoretical 
considerations.  For  the  practical  purposes  of  the  universe  in  which 
we  live  the  intrinsic  energy  may  be  considered  in  two  parts  ;  namely, 
available  and  unavailable  energy.  The  availability  of  the  intrinsic 
energy  of  a  body  depends  upon  external  conditions,  as  will  be  seen 
later  in  various  problems.  A  simple  mechanical  illustration  is  as 
follows.  Consider  the  velocity  of  a  cannon  ball  of  mass  m  to  be  t)02, 
then  its  energy  is  Jwt)02.  If  the  velocity  of  the  ball  after  it  has 
struck  a  ship  initially  at  rest  (that  is,  the  velocity  communicated 
to  the  ship  by  the  impact)  be  represented  by  t>,  then  the  intrinsic 
energy  of  the  ball  is  now  |  rat)2.  Of  the  total  energy  of  the  ball  only 
the  portion  |-m(t)02  —  t)2)  was  available  for  the  destruction  of  the 
ship. 


FUNDAMENTAL   LAWS  3 

3.  Molecular   Kinetic   Theory.     The   molecular   kinetic  theory  is 
based  upon  certain  assumptions  as  to  the  character  and  actions  of 
molecules  which  do  not  pretend  to  represent  the  actual  character 
of  the  molecules  but  which  simplify  the  problem  so  that  it  admits 
of  a  mathematical   analysis.     The  molecules   of   a  gas   (for  which 
state  many  of  the  assumptions  would  be  approximately  correct)  are 
considered  to  be  in  motion,  to  act  like  hard,  smooth,  small,  and  per- 
fectly elastic  spheres,  exerting  no  forces  upon  each  other  except  at 
the   instant  of  collision,  and  to  exist  in  almost  immeasurable  num- 
bers in  any  finite  volume  of  the  gas.     These  assumptions  admit  of 
the  application  of  the  laws  of  probability,  and  of  the  treatment  of 
collisions  as  if  the  molecules  were  uniform  spheres  of  unit  coefficient 
of  restitution,  colliding  centrally. 

Upon  these  assumptions  by  a  mathematical  analysis  Maxwell, 
Clausius,  Joule,  Boltzmann,  and  others  have  developed  conclusions  1 
as  to  the  behavior  of  a  gas  which  agree  with  the  observed  facts  as 
expressed  in  the  well-known  laws  of  Boyle,  Charles,  Dalton  (and 
Avogadro). 

As  expressed  by  Risteen,  one  of  these  conclusions  is  as  follows  : 
"  In  a  molecular  mixture  there  is  one  physical  quality  which  is  the 
same  for  every  set  of  molecules ;  and  that  is,  the  average  kinetic 
energy  of  translation  per  molecule.'''  Now  it  is  an  observed  fact  that 
whenever  two  or  more  substances  are  brought  into  intimate  contact 
there  is  one  physical  property  which  is  the  same  for  all  of  the  sub- 
stances, namely,  the  temperature.  Hence,  it  is  natural  to  assume 
that  equality  in  the  temperature  means  equality  in  energy  of  trans- 
lation of  the  molecules. 

4.  Temperature.     The  temperature  of  a  body  is  merely  an  arbi- 
trary expression  of  its  "  hotness  "  as  compared  with  some  states  of 
hotness  assumed  as  standard.     Because,  however,  of  the  inexactness 
of  the  sense  of  touch  as  a  measure  of  the  "hotness  "  and  because  of 
the  experimentally  determined  fact  that  some  substances  expand  as 
they  become  hotter,  the  expansion  or  contraction  of  a  given  amount 
of  such  a  substance  is  arbitrarily  taken  as  the  quantitative  measure 
of   a   temperature    change.     For   ordinary   purposes  the   substance 
used  is  mercury  enclosed  in  glass,  and  the  relative  expansion  of  the 
mercury  is  taken  as  a  measure  of  the  change  in  temperature. 

1  It  is  desirable  at  this  point  to  give  merely  the  briefest  possible  resume  of  the  kinetic 
theory  (the  facts  of  which  are  already  familiar  to  the  student  from  his  course  in  general 
physics),  in  order  that  the  concept  of  temperature  may  be  approached  from  the  side  of 
energy.  For  a  more  complete  statement  in  non-mathematical  language  see  Risteen,  "  Mole- 
cules and  the  Molecular  Theory,"  pp.  42-43. 


4  THERMODYNAMICS 

The  scale  upon  which  temperature  is  measured  has  been  chosen  in 
a  purely  arbitrary  manner.  On  the  Centigrade  scale  one  degree1 
of  temperature  represents  a  change  in  temperature  equal  to  that 
which  would  produce  one  one-hundredth  of  the  effect  produced  in 
the  mercury  thermometer  by  a  difference  in  temperature  corre- 
sponding to  the  following  easily  reproducible  states  of  pure  water. 
The  lower  point,  called  zero  degrees,  is  the  temperature  of  pure 
water  and  melting  ice  in  an  intimate  mixture.  The  upper  point, 
called  one  hundred  degrees,  is  the  temperature  of  the  vapor  of  pure 
boiling  water  in  contact  with  its  liquid.  In  both  cases  (although  it 
is  of  primary  importance  only  in  the  second  case)  the  pressure  upon 
the  mixture  must  be  what  is  known  as  standard  or  normal  pressure, 
which  is  equal  to  that  exerted  by  a  column  of  mercury  76  centi- 
meters high,  when  the  mercury  is  at  the  temperature  of  melting  ice 
and  water.  The  scale  is  extended  in  each  direction  from  these  two 
fixed  points  by  divisions  corresponding  to  a  degree  as  defined  above. 

The  Fahrenheit  scale  has  180  degrees  which  are  equivalent  to  100 
degrees  on  the  Centigrade  scale.  The  fixed  point  corresponding  to 
the  temperature  of  melting  ice  and  water  is  marked  32  degrees,  and 
the  boiling  point  is  212  degrees. 

According  to  the  molecular  kinetic  theory  two  bodies  are  at  the 
same  temperature  when  the  average  kinetic  energy  of  translation  of 
their  molecules  is  the  same.  It  will  be  shown  later  in  Section  15 
that  according  to  this  theory  one  degree  Centigrade  is  a  difference 
in  temperature  such  that  the  average  kinetic  energy  of  translation 
of  the  molecules  of  the  substance  considered,  is  increased  (or  di- 
minished) by  one  one-hundredth  of  the  change  in  this  kinetic  energy 
which  would  be  produced  by  a  difference  in  temperature  correspond- 
ing to  that  between  zero  and  one  hundred  degrees  Centigrade.  It 
will  also  be  seen  that  this  is,  for  a  gaseous  body,  approximately 
equal  to  ^y^  of  the  average  kinetic  energy  of  the  molecules  when 
the  substance  is  at  zero  degrees  Centigrade. 

5.  Thermal  Units.  When  two  bodies  at  different  temperatures 
are  placed  in  proximity,  it  is  found  that  they  ultimately  come  to  a 
common  temperature  (unless  this  difference  in  temperature  is  main- 
tained by  some  external  agency).  According  to  the  kinetic  theory 
this  must  mean  that  a  transfer  of  energy  has  taken  place  from  one 
body  to  the  other  in  order  that  the  final  average  molecular  energy 
of  translation  of  the  two  bodies  shall  be  the  same.  Energy  trans- 

1  For  more  rigorous  definition  of  a  degree  Centigrade  see  Section  14. 


FUNDAMENTAL   LAWS  5 

ferred  under  these  conditions  is  known  as  heat.  Because  it  was  not 
recognized  in  the  earlier  development  of  physical  science  that  heat 
is  energy,  units  were  introduced  for  the  measurement  of  heat. 

These  thermal  units  are  essentially  arbitrary  and  relative,  since 
they  depend  upon  the  unit  of  temperature  difference.  Thus  the 
heat  (energy)  required  to  change  the  temperature  of  one  gram  of 
pure  water  from  15°  to  16°  Centigrade  is  the  thermal  unit  in  the 
C.  G.  S.  system  and  is  called  the  calorie.  And  similarly  the  amount 
of  heat  (energy)  required  to  raise  the  temperature  of  one  pound  of 
water  from  62°  to  68°  Fahrenheit  is  called  a  British  thermal  unit,  or 
as  abbreviated  a  B.t.  u. 

6.  First  Law  of  Thermodynamics.     The  Principle  of  the  Conser- 
vation of  Energy  was  not  established   until    the  classical  work  of 
Joule  (and   Mayer)  performed   during   the    years  following    1843. 
This  principle  may  be  stated  in  a  more  general  form  than  that  given 
in  Section  2  as  follows  :    "  Every  physical   or  chemical    change  of 
state  has  a  fixed  mechanical  equivalent  expressible  in  work  units." 
Thus,  for  example,  the  amount  of  heat  (or  energy)  required  to  pro- 
duce in  water  the  change  in  physical  state  which  is  produced  by  a 
calorie  (or  a  B.  t.  u.)  is  to  be  equated  under  all  conditions  to  a  defi- 
nite number  of  work  units.     This  particular  application  of  the  Prin- 
ciple of  the  Conservation  of  Energy  was  experimentally  shown  to  be 
correct  by  Joule  and  was  really  the  basis  for  the  adoption  of  the 
principle  in  its  broadest  form.     This  application  has   received  the 
name  of  the  First  Law  of  Thermodynamics.     As  usually  stated  it 
reads :   "  Heat  and  mechanical  work  are  mutually  convertible,1  and 
heat  requires  for  its  production  and  produces  by  its  disappearance 
a  certain  definite   amount  of  work  for  each  thermal  unit."      The 
value  of  this  mechanical  equivalent  of  heat  from  the  determination 
made  by  Rowlands  is   778  foot   pounds  per  B.t. u.,  or  4.19  x  107 
ergs  per  calorie.     In  symbols  this  law  may  be  written 

TF=«flQorG  =  4TP; (1) 

where  Q  is  the  number  of  thermal  units,  TFthe  number  of  mechan- 
ical units  which  correspond,  and  J  is  the  (mechanical)  equivalent 
of  each  thermal  unit  in  work  units  for  the  system  in  which  W  is  to 
be  expressed.  The  factor  A  is  evidently  the  reciprocal  of  J. 

7.  Specific   Heat.      The  amount  of   heat  required    to  produce    a 
change  of   one  degree  in  the  temperature  of   a  substance  depends 

1  Except  in  so  far  as  all  the  heat  energy  of  a  hody  is  not  available  for  such  a  conversion, 
as  is  discussed  in  Section  2  and  also  in  later  sections. 


6  THERMODYNAMICS 

upon  the  nature  of  the  substance,  the  mass  in  which  it  is  desired  to 
produce  this  change,  and,  as  will  be  seen  in  Section  18,  upon  the 
conditions  under  which  the  change  is  produced.  For  the  same  mass 
of  a  given  Substance  it  is  found  that  the  amount  of  heat  required 
per  degree  is  different  at  different  temperatures  ;  that  is,  it  is  a 
function  of  the  temperature.  The  thermal  capacity  per  unit  mass 
or  "specific  heat  "  of  a  substance  is  denned  as  the  number  of  units  of 
heat  required  to  change  the  temperature  of  unit  mass  of  the  sub- 
stance one  degree.  This  is  expressed  either  in  calories  per  gram  or 
in  B.  t.  u.  per  pound,  and  the  numbers  representing  the  value  of  the 
specific  heat  of  a  given  substance  are  obviously  the  same  in  the  two 
systems.  But  it  would  be  better  to  express  the  specific  heat 
directly  in  mechanical  units,  since  these  belong  to  an  absolute  system 
and  since  this  may  readily  be  done  in  accordance  with  the  First 
Law  of  Thermodynamics.  Thus  J",  the  mechanical  equivalent  of 
heat,  is  merely  the  specific  heat  of  water  in  absolute  units. 

Since  the  specific  heat  of  a  substance  is  slightly  different  at  dif- 
fjerent  temperatures,  it  may  best  be  defined  by  a  differential  notation. 
Thus  if  a  quantity  of  heat  dc\  be  added  to  a  unit  mass  of  a  substance 
and  produce  an  increase  in  the  temperature  of  dt,  then  the  specific 
heat  <?,  at  the  temperature  £,  at  which  the  addition  was  made,  or  the 

heat  required  per  degree  per  unit  mass,  is  —  ^  =  c.     Hence  the  total 

dt 

heat  O  required  to  produce  in  a  mass  m  a  change  of  temperature 
from  £j  to  £2  is  /v 

Q  =  mjcd6  ........     (2) 

Frequently,  it  is  sufficiently  accurate  to  assume  the  specific  heat 
constant  through  the  range  of  temperature  ^  to  £2  and  to  write 


(3) 


where  c  is  the  average  value  of  the  specific  heat  of  the  substance  for 
a  range  of  temperature  which  includes  ^  and  £2. 

8.  Component  Parts  of  Specific  Heat.  When  the  temperature  of 
any  substance  is  changed  by  the  addition  of  a  quantity  of  heat,  it  is 
possible  to  consider  the  change  as  composed  of  three  parts,  and  it  is 
frequently  possible  to  assign  to  each  part  its  proper  portion  of  the 
heat  energy  transferred.  In  general,  heat  added  to  a  substance  may 
be  considered  as  producing  first,  a  change  in  the  temperature  ;  sec- 
ond, a  change  in  the  configuration  of  the  molecules  composing  the  sub- 
stance ;  and  third,  because  of  the  consequent  increase  (or  decrease) 


FUNDAMENTAL   LAWS  7 

of  the  volume  occupied  by  the  substance,  a  certain  amount  of 
external  work.  The  first  portion,  which  goes  to  produce  a  change 
in  the  temperature,  increases  according  to  the  molecular  theory  the 
average  kinetic  energy  of  translation  of  the  molecules.1  This  portion 
will  be  represented  by  the  symbol  AdS.  The  second  portion  of  the 
heat  supplied  produces  a  change  in  the  potential  energy  of  the 
molecules  and  will  be  represented  by  Adi.  The  third  portion, 
representing  an  amount  of  external  work,  will  be  symbolized  by  dW. 
In  general,  then,  if  a  small  amount  of  heat  c?Q  is  required  to  produce 
a  change  in  the  temperature  of  a  substance,  we  may  write 


(4) 


where  dS,  dl,  and  c?  IF  are  all  expressed  in  mechanical  units. 

In  the  case  of  the  change  of  state  from  water  to  steam,  to  be  dis- 
cussed later,  the  term  dl  is  large  as  compared  with  the  other  two 
terms.  In  the  case  of  a  "perfect  gas,"  also  to  be  discussed  later,  the 
term  dlis  zero  for  all  changes. 

9.  Latent  Heat.  In  the  case  of  a  change  of  state,  as  that  referred 
to  above,  where  water  is  converted  into  a  vapor,  or  vice  versa  (or  in 
the  case  of  the  formation  of  ice  from  water,  or  vice  versa)  there  is 
produced  no  change  in  temperature  until  sufficient  heat  energy  has 
been  added,  or  subtracted,  to  produce  the  change  in  potential  energy 
for  the  entire  mass  corresponding  to  the  different  configuration  of 
the  molecules.  During  such  a  change  of  state  the  term  AdS  is  then 

zer.o.     The  term  —  -,  representing  the  amount  of  heat  required  per 

Wit 

unit  mass  to  produce  the  change  in  internal  potential  energy,  is 
known  as  the  internal  latent  heat  for  that  particular  change  of  state. 
This  internal  latent  heat  is  also  a  function  of  the  temperature  at 
which  the  change  of  state  occurs,  as  will  be  discussed  more  fully 
later,  in  the  case  of  steam.  As  commonly  expressed  the  latent 

Adi         Ad  W 

heat  is  the  sum  of  the  terms  -  and  -  -  -,  the  latter  representing 

m  m 

the  external  latent  heat  or  external  work  which  must  be  done  in 
producing  the  change. 

Thus  in  C.  G.  S.  units,  the  latent  heat  of  vaporization  of  water  at 
100°  Centigrade  and  under  an  atmospheric  pressure  of  76  centi- 

1  It  is  true  also  that  an  increase  in  temperature  means  in  general  an  increase  in  the 
kinetic  energy  of  rotation  of  the  molecule  and  also  an  increase  in  the  "sub-molecular" 
kinetic  and  potential  energy  of  the  component  parts  of  the  molecule,  hut  for  the  further 
purposes  of  this  discussion  all  this  energy  will  be  considered  as  included  by  the  term  AdS. 


8  THERMODYNAMICS 

meters  of  mercury  is  536.5  calories.     Of  this  the  part  corresponding 

to  may  be  found  as  follows.     The  volume  occupied  bv  one 

m 

gram  of  water  changes  from  one  cubic  centimeter  to  1649  cubic 
centimeters,  which  is  the  volume  occupied  by  one  gram  of  water 
vapor  under  the  given  condition  of  temperature.  Since  this  change 
takes  place  against  a  pressure  of  (76)  (13.6)  (980)  dynes  per  square 
centimeter,  (76)(13.6)(980)  (1649-1),  or  1.668  x  109,  ergs  repre- 
sent the  external  work  during  evaporation.  That  is,  the  external 
latent  heat  of  evaporation  is  1.668  x  109  -r-  4.19  x  107  =  40  calories 
for  the  conditions  of  the  problem.  The  internal  latent  heat,  repre- 
senting the  gain  in  molecular  potential  energy,  is  then  496.5  calories 
or  2.08  x  1010  ergs. 

10.  Boyle's  Law.  Boyle's  Law  states  that,  other  things  being 
equal  (that  is,  the  temperature,  mass,  and  molecular  constitution  of 
the  gas  experimented  upon  remaining  unchanged),  the  product  of 
the  pressure  at  which  a  given  gas  is  maintained  and  the  volume 
which  it  occupies  is  a  constant.  Or,  in  other  words,  if  the  pressure 
is  changed,  the  volume  occupied  by  a  given  quantity  of  gas  main- 
tained at  a  constant  temperature  undergoes  a  change  inversely  pro- 
portional to  the  change  in  pressure.  In  symbols  this  is  expressed  as 

pV=  a  constant,  or  Tr          rr          T_  >-* 

Pi^i  =Pz  F2  =Ps  V» O) 

where  p  represents  the  pressure  or  force  per  unit  area  and  ^repre- 
sents the  volume  (usually  measured  in  the  same  system  of  units 
as  is  the  area). 

The  value  of  the  constant  is  evidently  for  a  given  mass  of  some 
particular  gas  a  function  of  the  temperature.  This  follows  at  once 
from  the  observable  fact  that  if  the  temperature  is  changed,  the  value 
of  the  product  p  V  is  also  changed.  The  physical  meaning  of  this 
fact  is  evident  from  the  following  analysis  based  upon  the  assump- 
tions of  the  kinetic  theory. 

The  pressure  upon  the  bounding  surface  of  a  body  of  gas  is  due 
to  the  impacts  of  the  moving  molecules.  It  is  proportional  to  the 
number  of  these  impacts  per  second  upon  each  unit  area  of  the 
bounding  surface,  and  to  the  impulse  of  each  molecular  impact. 
The  impulse  is  proportional  to  the  change  in  momentum  of  the  mov- 
ing molecule,  and  this  in  turn  to  the  velocity  and  mass  of  the  mole- 
cule. The  total  number  of  impacts  made  by  a  given  molecule  will 
be  proportional  to  its  velocity.  The  number  of  molecular  impacts 
upon  each  unit  of  area  of  the  bounding  surface  will  be  proportional 


FUNDAMENTAL   LAWS  9 

to  the  gas  density  ;  that  is,  for  a  given  mass  of  gas,  inversely  propor- 
tional to  the  entire  volume.  Hence  the  pressure  is  directly  propor- 
tional to  the  mass  of  a  molecule  and  to  the  square  of  the  average 
molecular  velocity,  and  inversely  proportional  to  the  entire  volume. 
That  is,  the  product  of  the  pressure  and  the  volume  is  directly 
proportional  to  the  average  molecular  kinetic  energy  of  translation, 
and  hence  is  some  function  of  the  temperature.1 

This  law  of  Boyle  is  not  rigorously  correct,  for  all  gases  show 
departures  from  it  at  high  pressures.  It  is  followed  very  closely  for 
pressures  under  10  atmospheres.  For  higher  pressures  the  de- 
partures are  more  marked.  Thus  for  air  they  range  from  about 
^  per  cent  at  10  atmospheres  to  as  much  as  25  per  cent  at  600 
atmospheres. 

11.  Joule* s  Law.  Another  important  law  of  gases  is  due  to 
experiments  performed  by  Joule.  This  law  states  that  "  all  the 
work  done  in  compressing  a  gas  is  conducted  away  as  heat  if  the 
temperature  of  the  gas  is  kept  constant,  and  conversely,  when  a  gas 
expands  at  constant  temperature,  it  must  receive  from  its  surround- 
ings a  quantity  of  heat  equal  to  the  work  done  by  the  expansion." 
This  means  that  during  the  expansion  of  a  gas  no  internal  work  is 
done  by  the  molecules  either  by  molecular  repulsions  or  against 
molecular  attractions.  For  if  there  are  attractive  forces  between 
the  molecules,  then  in  such  an  expansion,  since  the  molecules  become 
more  widely  separated,  work  must  be  done  to  produce  this  separa- 
tion, and  this  must  be  supplied  by  the  partial  conversion  of  molec- 
ular kinetic  energy  into  molecular  potential  energy.  If  '  there 
is  to  be  no  consequent  lowering  in  the  temperature,  there  must  be 
supplied  to  the  molecules  of  the  substance  from  the  outside  an 
amount  of  heat  energy  equal  to  that  converted  into  molecular  poten- 
tial energy.  The  amount  of  heat  supplied  to  an  expanding  gas 
would  then  be  greater  than  the  amount  of  external  work  performed 
by  the  gas  during  expansion  by  this  amount  of  increase  in  the 
molecular  potential  energy.  According  to  Joule's  Law  there  are  then 
no  molecular  attractive  forces  acting  between  molecules.  On  the  other 
hand,  if  repellent  forces  exist  between  the  molecules,  the  potential 
energy  term  will  decrease  in  value,  and  there  will  be  a  larger  amount 
of  external  work  done  than  there  is  heat  supplied  to  the  expanding 
gas.  Thus,  the  possibility  of  repellent  forces  between  the  mole- 
cules of  a  gas  is  also  denied  by  Joule's  Law.  In  so  far,  then,  as  a 

1  See  any  general  physics  text,  or  Millikan,  "  Mechanics,  Molecular  Physics,  and  Heat," 
Chapters  XIV  and  XVI. 


10 


THERMODYNAMICS 


.  -** 

*  —  >  —  - 

"  —  N 

\ 

,' 

B 

A 

c 

\\*~  ""Yl 

T-I'~ 

.)}    IL_ 

..::-{' 

FIG.  1 


gas  obeys  Joule's  Law  the  potential  energy  term  dl  in  the  equation 
d&  =  A(dS  +  dl  -\-  dW)  must  be  zero.  This  law  is  not  more  rigor- 
ous than  Boyle's  Law,  and  the  departures  from  Boyle's  Law  may  be 
explained  in  part  upon  the  assumption  that  there  are  molecular 
attractions  which  are  not  entirely  negligible  when  the  density  is 
high  and  the  molecules  close  together. 

Joule's  classical  experiment  was  performed  with  apparatus  shown 
diagrammatically  in  Figure  1.  Two  similar  metal  vessels  A  and  B, 

connecting  by  a  tube  fitted  with 
a  stopcock  C,  were  enclosed  in  a 
large  calorimeter.  The  tempera- 
ture was  measured  by  a  ther- 
mometer reading  to  0°.005  Fahren- 
heit.1 The  vessel  A  contained  air 
under  a  pressure  of  about  22  at- 
mospheres, and  the  vessel  B  was 
practically  a  vacuum.  When  the 
stopcock  was  opened,  the  air  in  A 
expanded  into  B  without  doing 
external  work.  It  was  found  im- 
possible to  detect  any  change  in  the  temperature  of  the  water  as  a 
result  of  this  expansion.  From  this  experiment  Joule  concluded 
that  "  no  change  in  temperature  occurs  when  air  is  allowed  to  ex- 
pand in  s.uch  a  manner  as  not  to  develop  mechanical  work."  From 
this  there  follows  the  conclusions  stated  above  as  to  the  intrinsic 
potential  energy  of  the  molecules  of  a  gas. 

12.  Perfect  Gases.  Since  all  gases  show  departures  from  the  ap- 
proximate laws  of  Boyle  and  Joule,  but  since  these  departures  are 
not  in  general  large,  it  is  convenient  for  the  purposes  of  analysis  in 
problems  dealing  with  gases  to  treat  of  purely  fictitious  so-called 
"perfect  gases."  A  perfect  gas  is  defined  as  one  for  which  the  laws 
of  Boyle  and  Joule  hold  rigorously  for  all  pressures  and  tempera- 
tures. Such  a  gas  would  then  have  no  interacting  forces  between 
its  molecules  except  at  the  instant  of  mutual  impact,  and  according 

1  The  more  accurate  "porous  plug  experiment"  of  Kelvin  and  Joule  is  described  later 
(see  Section  35),  but  merely  as  an  illustration  of  an  entropy  change  in  an  irreversible  adia- 
batic  expansion.  The  conclusions  of  that  experiment  which  bear  upon  this  discussion  are 
here  stated.  At  ordinary  temperatures  there  is  a  small  but  appreciable  attraction  between 
the  molecules  of  a  gas,  and  hence  a  cooling  by  free  expansion  through  an  orifice.  The  cool- 
ing is  roughly  0°.25  Centigrade  per  atmosphere  of  pressure  difference  on  the  two  sides  of  the 
orifice.  The  value  is  less  for  higher  temperatures,  and  greater  for  lower  temperatures. 
This  phenomenon,  successively  caused  to  take  place  in  the  same  mass  of  gas,  is  of  course 
the  basis  of  liquid  air  machines. 


FUNDAMENTAL   LAWS  11 

to  the  kinetic  theory  (see  Section  10,  above)  the  product  of  the 
pressure  and  the  volume  would  be  some  function  of  the  temperature. 
Writing  p  V=f(t),  where  f(t)  represents  some  function  of  the 
temperature,  it  is  possible  to  show  that  for  a  perfect  gas  the  values 
of  the  "pressure  coefficient"  and  of  the  "volume  coefficient"  are 
equal  when  determined'  for  the  same  temperature.  The  pressure 
coefficient  /3  is  defined  as  the  increase  in  pressure  per  degree  of 
temperature  per  unit  of  pressure.  In  symbols  this  is  expressed  as 


where  pQ  is  the  original  pressure  exerted  by  a  given  mass  of  gas  at  a 
temperature  of  £0,  .and  p^  and  t^  represent  the  pressure  and  tempera- 
ture of  some  later  condition  of  the  gas  while  occupying  the  same 
volume.  In  differential  notation  this  is  most  conveniently  expressed 
by  using  the  symbol  (  )F  to  indicate  the  fact  that  the  quantity 
enclosed  by  the  parenthesis  is  found  under  the  conditions  of  constant 
volume.  Thus 


Similarly,  the  expansion  coefficient  a  is 


From  the  expression  of  Boyle's  Law  for  a  perfect  gas  as  p  V=f(t) 

wehave 


P\Vdtv 


_ 
" 


V\  dt  )P     pV     /CO 

The  values  of  a  and  of  (3  are  evidently  equal  for  the  same  values  of 
the  temperature  t. 

The  fact  that  all  gases  have  the  same  value  of  a  for  any  given 
temperature  was  discovered  as  an  experimental  fact  by  Charles  in 
1787,  some  sixty  years  before  the  announcement  of  Joule's  Law, 
which  has  been  used  here  in  the  definition  of  a  perfect  gas. 

13.  Charles's  Law.  Charles's  Law  states  that  all  gases  show  the 
same  fractional  or  per  cent  expansions  between  the  limits  of  two 
fixed  temperatures.  Later  experiments  have  shown  that  this  law 
of  Charles  (or  of  Gay-Lussac,  as  it  is  sometimes  known)  is  only 


12  THERMODYNAMICS 

approximate.  Experiments  have  also  shown  that  for  all  gases  the 
values  of  the  pressure  coefficient  /3  for  the  same  initial  temperature 
are  practically  equal,  and  are  also  practically  equivalent  in  value  to 
the  values  of  the  expansion  coefficient  a  for  the  same  temperature. 
This  is  evident  from  Table  I,1  where  are  given  for  various  gases 
values  of  a  and  /3  which  represent  their  average  values  for  the  range 
of  temperature  from  0°  to  100°  Centigrade. 

The  definition  of  a  perfect  gas  leads  then  to  the  conclusion  that  for 
perfect  gases  Charles's  Law  holds  rigorously,  and  also  the  values  of 
a  and  /?  are  identical.  For  a  perfect  gas  the  value  of  /3  will  be  taken 
as  0.0036625  at  0°  Centigrade,  as  is  explained  in  Section  15. 

14.  Standard  Hydrogen  Thermometer.     Because  of  the  fact  that 
the  expansion  of  most  solids  and  liquids  is  not  a  linear  function  of 
the  temperature,  while  that  of  a  gas  is  very  nearly  a  linear  function 
in  accordance  with  Charles's  Law,  the  standard  thermometric  sub- 
stance is  a  gas.     A  gas  thermometer  might,  obviously,  with  almost 
equal  justification,  be  constructed  to  measure  temperature  by  the 
expansion  of  the  gas  under  the  condition  of  constant  pressure,  or  by 
the  change   in  pressure   under  the   condition   of  constant  volume. 
Because  of  the   experimental   difficulty  of  making  the   corrections 
which  enter  into  a  determination  of  temperature  with  a  constant 
pressure  gas   thermometer   (due  to  the  variations  in  the  exposed 
length  of  the  column  of  gas  contained  in  the  stem  of  the  thermome- 
ter),  Regnault  adopted  the  constant  volume  form.     In  1887  the 
International   Committee  of   Weights   and    Measures   adopted   the 
constant  volume  hydrogen  thermometer  as  the  standard.     According 
to  this  agreement  one  degree   Centigrade  is  such  a  temperature  as 
will  cause  a  change  in  the  pressure  of  a  mass  of  hydrogen  (kept  at 
a  constant  volume)  of  0.0036625,  or  ^3-,  of  the  pressure  which  this 
hydrogen  exerts  at  the  temperature  of  melting  ice  and  pure  water. 

15.  Absolute  Temperature.     According  to  the  discussion  of  Sec- 
tion 10  for  a  constant  volume  perfect  gas  thermometer  we  may 
write  the  pressure  as  proportional  to  the  average  molecular  kinetic 
energy   of    translation.     If,    further,  the  pressure  coefficient  $  be 
taken  for  a  perfect  gas  as  equal  to  the  value  0.0036625,  whch  is  given 
by  Chappuis  for  hydrogen,  a  gas  which  most  nearly  obeys  the  laws 
of  perfect  gases,  it  may  be  shown,  as  follows,  that  one  degree  Centi- 
grade change  in  the  temperature  of  a  perfect  gas  corresponds  to  a 
change  in  the  average  molecular  kinetic  energy  of  273-  part  of  the 

i  See  page  130. 


FUNDAMENTAL  LAWS  13 


value  at  zero  degrees  Centigrade  of  this  average  molecular  kinetic 
ergy. 

Hence 


energy.     Thus  T, 

.  K.  E. 


dt 
and 


(dp\       / 
\&)T\ 


p\dt)v     av.  K.  E.V          dt          Jv 
or 


1    _  1  A?(av.  K.  E.)\ 

av.  K.  E.atO°C.V          dt          )} 


273      av.  K.  E.atO°C.V          dt  Jratooc.,      •      •     (8) 


where  the  symbol  "  (  )Fat  0°  C."  is  used  to  represent  the  fact  that 
the  volume  is  maintained  constant  at  the  value  which  it  has  for  a 
temperature  of  zero  degrees  Centigrade. 

If,  then,  the  temperature  of  a  perfect  gas  be  lowered  to  —  273° 
Centigrade,  it  follows  that  the  molecules  would  be  completely  de- 
prived of  kinetic  energy  of  translation,  and  hence  at  rest.  The  pres- 
sure exerted  by  the  gas  would  then  be  zero.  This  temperature  of 
—  273°  Centigrade  is  -known  as  the  absolute  zero;  for,  if  temperatures 
are  measured  in  degrees  absolute,  that  is  in  degrees  from  this  abso- 
lute zero,  the  intrinsic  energy  of  a  perfect  gas  is  directly  proportional 
to  its  temperature.  For  all  practical  problems  the  intrinsic  energy 
of  the  so-called  permanent  gases  may  also  be  taken  to  be  propor- 
tional to  the  absolute  temperature. 

This  absolute  zero  is  not  an  attainable  temperature.  All  known 
gases  liquefy  and  even  solidify  above  this  temperature.  Thus  hydro- 
gen liquefies  at  about  -  252°  C.  and  solidifies  at  about  -  256°  C. 
By  evaporating  liquid  helium  a  temperature  of  about  —  270°. 5  C.  has 
been  attained.  These  low  temperatures  are  measured  with  a  plati- 
num resistance  thermometer  which  has  been  found  to  give  results 
very  close  to  that  of  a  hydrogen  thermometer. 

For  a  perfect  gas,  and  for  most  practical  problems  in  gases,  under 
the  conditions  of  constant  volume  the  pressure  may  be  taken  pro- 
portional to  the  absolute  temperature.  This  is  expressed  in  symbols 
in  equation  (9).  And  similarly,  for  constant  pressure,  the  volume 
may  be  taken  proportional  to  the  absolute  temperature,  as  is  expressed 
in  symbols  in  equation  (10) 

.     \  //3    \ 

(9) 

.......     (10) 


14  THERMODYNAMICS 

where  6  is  the  absolute  temperature  as  measured  on  the  perfect  gas 
scale.  Or,  in  symbols, 

00  =  273.     0l  =  tl  +  273  in  degrees  Centigrade (11) 

And  since  273°  Centigrade  equals  491°. 5  Fahrenheit, 

00  =  459.5.     01  =  fj  +  459. 5  in  degrees  Fahrenheit. .     .     .     (12) 

16.  General  Equation  for  a  Perfect  Gas.     It  is  evident  since 

p  Foe  average  molecular  K.  E.  oc#, 

that  a  general  expression  for  the  pressure,  volume,  and  absolute 
temperature  of  a  perfect  gas  may  be  written  as  follows: 

pV  =  mRO, (13) 

where  j?  is  pressure,  Fis  volume,  0  is  the  absolute  temperature,  m  is 
the  mass  of  the  gas,  and  R  is  a  constant.  The  value  of  R  depends 
upon  the  kind  of  gas  and  has  a  numerical  value  which  represents  in 
the  system  of  units  to  which  m  belongs  the  value  of  the  expression 

^  for  unit  mass  of  the  gas  for  any  condition  for  which  the  corre- 
ct 

sponding  values  of  p,  v,  and  0  are  known. 

Thus  for  air,  assuming  that  it  will  follow  the  equation  for  a  per- 
fect gas,  and  finding  R  for  the  C.  G.  S.  system,  we  have  the  density 
of  air  at  the  standard  conditions  of  76  centimeters  of  mercury  pres- 
sure and  zero  degrees  Centigrade  equal  to  0.001293  as  given  by 
Regnault  for  latitude  45°  (where  g  is  980.6).  The  volume  of  one 
gram  of  air  for  these  conditions  is  then  773.4  cubic  centimeters. 
The  value  of  p  is  1.0134  x  106  dynes  per  square  centimeter.  The 
absolute  temperature,  0,  is  273.  Hence  R  is  2.867  x  106. 

Similarly  for  the  F.  P.  S.  system,  the  volume  of  one  pound  of  air 
is  12.39  cubic  feet  at  a  pressure  of  14.7  pounds  per  square  inch  and 
at  a  temperature  of  32°  Fahrenheit.  Expressing  the  pressure  p  as 
144  x  14.7  pounds  per  square  foot  and  substituting,  gives  R  equal 
to  53.35. 

17.  Use  of  the  Term  "Specific."     As  used  in  physics  the  meaning 
of  the  term  "  specific "  may  best  be  shown  by  illustration.     Thus 
"  specific  heat "  is  the  amount  of  heat  required  to  produce  a  definite 
result,  namely,  unit  change  of  temperature  in  unit  mass.     The  term 
"specific"  then  indicates  on  the  part  of  the  physical  magnitude  to 
which  it  is  applied  a  definite  and  peculiar  quality  in  which  this  par- 
ticular value  of  the  magnitude  differs  from  all  other  possible  values 
of  the  same  magnitude.     In  general  it  limits  the  magnitude  to  unit 


FUNDAMENTAL   LAWS  15 

mass,  unit  length,  unit  area,  unit  volume,  or  to  a  unit  change  in 
some  related  magnitude.  Thus  "  specific  resistance  "  is  ths  resist- 
ance per  unit  cube,  or  the  resistance  per  unit  length  having  unit 
cross  section,  as  in  the  case  of  the  mil-foot. 

The  adjective  "  specific  "  is  somewhat  more  widely  used  in  engi- 
neering, where  the  terminology  is  not  otherwise  as  carefully  differ- 
entiated as  in  pure  physics.  Thus  in  the  rigorous  physical  usage 
"  pressure  "  always  means  "  force  per  unit  area,"  but  in  engineering 
usage,  broadened  and  rendered  less  precise  by  the  influence  of  the 
necessity  of  conveying  ideas  in  popular  language,  pressure  and  force 
are  used  almost  interchangeably.  For  this  reason  there  has  come 
into  common  engineering  use  the  term  "  specific  pressure  "  to  mean 
force  per  unit  area,  as  for  example  in  the  F.  P.  S.  system,  where  spe- 
cific pressure  is  the  force  in  pounds  per  square  foot.  Similarly, 
"density,"  which  to  the  student  of  physics  means  "mass  per  unit 
volume,"  is  expressed  by  the  engineer  as  "  specific  weight,"  meaning 
thereby  the  weight  per  cubic  foot.  "  Specific  volume  "  is  the  volume 
of  one  pound  in  cubic  feet.  "  Specific  gravity  "  is  the  weight  of  a 
given  volume  of  a  substance  in  terms  of  the  weight  of  the  same 
volume  of  water,  taken  as  unity.1 

For  the  purposes  of  this  book  the  word  "pressure"  will  always  be 
used  in  its  rigorous  sense  of  force  per  unit  area.  Also,  the  follow- 
ing scheme  of  notation  has  been  adopted  to  avoid  confusion.  When- 
ever volume,  quantity  of  heat,  or  any  other  unit  which  may  later  be 
introduced  refers  to  unit  mass  of  the  substance  under  consideration, 
a  lower  case  letter  will  be  used.  Whenever  reference  is  made  to 
the  total  value  of  a  magnitude,  referring  to  some  other  mass  than 
unity,  a  capital  letter  will  be  used.  It  will  be  noticed  that  this  has 
been  done  consistently  in  the  earlier  portion  of  the  text;  thus, 
p  V=  mR0,  but  pv  =  RO. 

18.  Specific  Heat  of  Gases.  The  specific  heat  of  a  gas  may  be 
determined  under  the  condition  either  of  constant  volume  or  of 
constant  pressure.  It  is  evident  from  a  consideration  of  equation 
(4)  of  Section  8  that  when  the  volume  is  constant,  since  the  term 
dW  is  zero  (because  there  is  no  external  work  done  in  expanding), 

1  This  inexactitude  of  common  engineering  terminology  has  frequent  illustration.  Thus 
"speed"  and  "velocity"  are  used  interchangeably.  "Acceleration,"  which  is  the  rate  of 
change  of  velocity,  is  popularized  into  "  rate  of  acceleration,"  an  expression  which  rigor- 
ously is  a  second  derivative  of  velocity  instead  of  the  first.  ''Electromotive  force"  and 
"potential  difference  "  are  frequently  used  as  synonymous.  Also  "  energy  "  and  "  power  " 
are  not  carefully  distinguished,  as  is  illustrated  by  the  frequent  use  of  expressions  like 
"  expenditure  of  power." 


16  THERMODYNAMICS 

the  amount  of  heat  required  to  produce  in  unit  mass  a  temperature 
change  of  one  degree  is  less  than  that  required  to  produce  this  change 
when  the  pressure  is  kept  constant  and  the  term  dWh&s  a  numeri- 
cal value  (representing  the  work  done  externally  by  the  expansion 
of  the  gas  against  the  constant  pressure  to  which  it  is  subjected). 
These  two  values  of  the  specific  heat  of  a  gas  will  in  the  future  be 
represented  by  cp  and  cv  for  the  conditions  of  constant  pressure  and 
constant  volume,  respectively.  Or,  in  symbols, 


The  ratio  of   the  specific  heats  of  a  gas     written  K  =  -^  )may  be 

V  cvj 

shown  to  depend  upon  the  atomic  structure1  of  the  gas  considered. 
For  monatomic  gases  this  ratio  K  has  a  value  of  1.66,  for  diatomic 
gases  a  value  of  about  1.40;  for  the  diatomic  mixture  constituting 
air  the  value  is  1.401,  for  carbon  dioxide  it  is  1.30  (at  30°  C.). 

19.  Graphical  Representation  of  Gas  Transformations.  Changes  in 
the  values  of  the  magnitudes  p,  V,  and  6  for  a  given  mass  of  gas  may 
best  be  represented  graphically  by  plotting  on  a  set  of  coordinate 
axes  values  of  V  as  abscissas  and  values  of  p  as  ordinates.  Since 
the  temperature  6  is  a  function  of  p  and  F",  it  is  implicitly  represented 
upon  the  same  plot. 

The  possible  ways  in  which  a  given  mass  of  gas  may  undergo  a 
continuous  transformation,,  not  involving  any  chemical  change,  are 
evidently  the  following  :  (#)  at  constant  volume  by  changing  the 
pressure  indirectly  through  a  change  in  the  temperature  ;  (5)  at  con- 
stant pressure  by  changing  the  volume  indirectly  through  a  change 
in  the  temperature  ;  (<?)  at  constant  temperature,  that  is  isothermally  , 
and  hence  in  accordance  with  the  relation  of  Boyle's  Law;  the  values 
of  p  and  T^are  then  related  as  p  V=pl  V^  =  a  constant  ;  (d\  without 
the  addition  or  subtraction  of  heat,  that  is,  as  it  is  called,  adiabati- 
cally  ;  as  will  be  seen  in  Section  21  the  equation  expressing  this 
relation  is  pV*  =  PI  VK^  =  a  constant;  (0)  in  accordance  with  the 
condition  that  the  intrinsic  energy  shall  be  constant,  that  is,  an 
isoenergic  transformation  ;  for  a  perfect  gas  this  condition  is  evidently 
equivalent  to  that  of  an  isothermal  change  ;  (/)  the  change  may  be 
under  the  condition  of  constancy  in  what  is  called  the  entropy  of 
the  gas,  that  is,  the  change  may  be  isentropic  ;  for  a  reversible  trans- 
formation, as  will  be  discussed  in  Section  31,  this  is  equivalent  to 

i  See  Edser,  "  Heat  for  Advanced  Students,"  p.  302. 


FUNDAMENTAL   LAWS 


17 


an  adiabatic  change  ;  or  (#)  the  transformation  may  not  be  in  ac- 
cord with  any  of  the  above  conditions,  in  which  case  frequently  it 
may  be  represented  by  a  relation  of  the  form  p  Vn  =  p1  V-f  =  a  con- 
stant ;  or,  it  may  be  assumed  to  take  place  in  a  series  of  infinitesi- 
mal steps  which  are  in  accordance  with  some  of  the  above  conditions. 
The  plots  of  Figure  2,  lettered  #,  6,  c,  c£,  and  g,  represent  graphi- 
cally changes  in  accordance  with  the  conditions  of  a,  6,  <?,  c?,  and  g^ 


V, 


V,  V2 

6 


V,  V5 


V,  V, 


V2 


c 
FIG.  2 


above,  from  a  volume  and  pressure  of  V1  and  p1  to  a  volume  and 
pressure  of  Vz  and  p2,  respectively.  The  plot  (#)  is  drawn  to  a  very 
much  larger  scale  than  the  others  and  represents  disproportionately 
a  change  as  taking  place  in  two  imaginary  or  component  infinitesi- 
mal steps;  namely,  from  (pv  V^)  to  (p.2,  F3)  by  an  adiabatic  com- 
pression and  from  (j?2,  F3)  to  (^  V^)  by  an  isothermal  expansion. 

20.  Work  Calculations  for  Gas  Transformations.  The  work  re- 
quired to  produce  a  change  in  the  state  of  a  gas,  or  the  external 
work  done  by  the  gas,  during  some  transformation,  may  be  repre- 
sented graphically  upon  the  pressure-volume  plot  just  described. 
Thus,  consider  that  a  quantity  of  gas  is  contained  in  a  vessel  fitted 
with  a  piston  of  area  a.  If  the  pressure  of  the  gas  is  p  units  of  force 
per  unit  of  area,  then  the  total  force  exerted  upon  this  piston  is 
F—  pa  units.  If  the  piston  is  caused  to  move  a  small  distance  dz 
by  this  force  (or  against  it),  the  infinitesimal  amount  of  work  dW 
is  dW=  Fdz  =  padz.  During  this  motion  the  volume  is  changed  by 
an  infinitesimal  amount  dV,  which  is  evidently  equal  to  adz.  The 
expression  for  the  work  done  in  producing  this  change  is  then 
dW=Fdz=padz  =  pdV.  This  may  be  considered  positive  if  the 
gas,  or  working  substance,  produces  the  change  and  does  this  amount 


18  THERMODYNAMICS 

of  external  work,  and  negative  if  the  work  is  done  by  an  external 
agent  upon  the  working  substance.  The  total  work  required  to 
produce  the  change  from  a  volume  V^  to  a  volume  V^  is  then 


(15) 


If  the  summation  is  performed  between  the  limits  as  shown  above, 
it  is  evident  that  an  expansion  will  be  positive  and  correspond  to 
work  done  by  the  substance,  and  similarly  a  compression  will  be 
negative  and  correspond  to  work  done  upon  the  substance. 

The  integration  given  above  is  only  possible  analytically  when  p  is 
either  constant  or  some  known  function  of  F".  In  the  "jo-Fplot," 
however,  the  work  required  to  produce  a  given  change  in  volume  is 
always  to  be  represented  by  the  area  enclosed  between  the  axis  of 
volumes,  the  curve  p  =/(F"),  and  the  two  ordinates  of  values  Vl  and 

rf 

It  is  important  to  note  that  although  the  above  discussion  referred 
specifically  to  gases,  the  integral  \  pd  V  between  the  proper  limits 

always  gives  the  mechanical  work  required  to  produce  the  change  in 
volume  indicated.  In  the  case  of  the  transformations  of  steam  in 
a  steam  engine  the  pressure-volume  plot  is  drawn  by  a  mechanism 
known  as  an  indicator,  which  is  attached  to  the  cylinder  of  the 
engine  in  such  a  manner  as  to  record  pressures  and  to  the  piston  of 
the  engine  so  as  to  record  piston  displacements.  The  work  done  by 
the  steam  is  then  found  approximately  by  using  a  planimeter  to 
determine  the  area. 

The  cases  mentioned  in  Section  19  and  shown  in  Figure  2,  a,  b,  c, 
and  c?,  will  now  be  discussed.  For  completeness,  in  addition  to 
representing  the  value  of  the  work  integral  there  will  be  recorded 
at  this  time  the  energy  relations  and  the  heat  relations  for  these 
changes,  although  some  of  these  relations  are  not  fully  developed 
until  Sections  21  and  22.  The  following  symbols  will  be  used :  to 
represent  intrinsic  energy  .27;  to  Represent  heat  added  to  the  sub- 
stance from  some  external  source  £}.  The  subscripts  1  and  2  will 
represent  states  1  and  2  respectively.  It  is  to  be  noticed  that  the 
relations  given  below  hold  only  for  a  perfect  gas  or  one  which  may 
be  assumed  perfect,  and  in  every  instance  of  a  change  of  volume  the 
change  is  taken  as  an  expansion ;  that  is,  F^  is  greater  than  Vr 

(#)  Since  the  volume  is  constant,  W=  0.  Because  the  tempera- 
ture is  increased  the  intrinsic  energy  is  increased  and  by  an  amount 

fl.«(^-<1) (16) 


FUNDAMENTAL   LAWS  19 

Since  p  is  constant, 

-V^).     Also 


EI-EI  .......    (17) 

For  an  isothermal  expansion 

U2  --#!  =  <),  and  Q  =  A  W,  where 


oge!.       •     (18) 

^2 

(6?)  For  an  adiabatic  expansion  Q  =  0,  and  W=  U2  —  E^  where 


<"> 


These  integrations  should  be  verified  by  the  student.  The  results 
will  be  used  in  subsequent  discussions. 

21.  Adiabatic  Transformation.  The  equation  stated  in  Section  19 
as  representing  the  relation  for  an  adiabatic  transformation  of  a  gas 
will  now  be  deduced.  Consider  a  temperature  change  of  infinitesi- 
mal amount  taking  place  in  a  unit  mass  of  the  gas  as  the  result  of 
the  addition  of  an  infinitesimal  amount  of  heat  energy  £q.  Imagine 
the  change  to  take  place  in  two  infinitesimal  steps  as  follows,  a 
change  in  temperature  of  amount  (£#)„  while  the  volume  remains 
constant,  and  a  change  of  (&@)p  while  the  pressure  remains  con- 
stant. Since  part  of  this  change  is  at  constant  volume  and  part  at 
constant  pressure,  the  amount  of  heat  Sq  required  to  produce  the 
change  is  the  sum  of  the  heat  required  to  produce  these  component 
changes,  namely,  cv(§6}v  and  cp(§6)p,  where  cv  and  cp  are  the  specific 
heats  at  constant  volume  and  at  constant  pressure  respectively. 

Thus  *H=cv(*0\  +  cp<$e\.        ....     (20) 

Since,  however,  for  this  unit  mass  of  gas  p  and  v  are  related  by  the 
equation  pv  =  R6,  it  follows  that  a  change  of  temperature  of  amount 
($#),  corresponds  to  and  is  expressible  by  (v/R)8p,  and  similarly 
a  change  of  temperature  (80)  p  is  expressible  as  (p/R)8v.  Making 


20  THERMODYNAMICS 

this  change  of  variables  and  writing  in  differential  notation  instead 
of  the  notation  of  infinitesimals,  gives 


(21) 


Equation  (21)  is  a  general  heat  equation  for  a  perfect  gas  and 
expresses  the  effect  of  adding  heat  energy  in  terms  of  the  changes 
of  the  pressure  and  the  volume  of  the  gas.  In  Section  38  similar 
equations  will  be  written  expressing  this  effect  in  terms  of  the  other 
possible  variables  of  the  perfect  gas  equation,  namely,  p  and  0,  and 
v  and  0. 

In  the  case  of  an  adiabatic  transformation,  that  is  by  definition 
one  which  takes  place  under  conditions  which  prevent  the  transfer 
of  heat  to  or  from  the  substance,  obviously  the  term  dc\  is  zero,  and 
equation  (21)  becomes 

c,^dp  +  cp^dv  =  0  .......     (i) 

Upon  substitution  of  K  =  cp/cv  (see  d3fmitioii  of  K  in  Section  18), 

this  gives  K^L  =  -clP.          .....     (ii) 

v  p 

Solving  this  differential  equation  by  integration  between  the  limits 
of  pl  and  vr  which  represents  the  condition  of  the  gas  in  the  first 
state,  and  p2  and  va,  which  represents  its  second  condition,  gives 

log/^Y  =  log,  £1,     ......     (iii) 

\«V  p2 

whence  pv  "  =  p^vf  =  p<p£,        ....      (iv) 

or  using  the  volume  of  m  units  instead  of  one  unit  of  mass, 

PJ\"=P*V{'  (22) 

22.  Comparison  of  Adiabatic  and  Isothermal  Transformations.  It 
is  evident  from  equation  (22)  that  the  curve  representing  an  adia- 
batic transformation  on  a  pressure-volume  plot  is  steeper  than 
that  representing  an  isothermal  transformation,  inasmuch  as  the  tan- 
gent of  the  angle  of  slope,  given  by  dp/dv  is  K  times  as  large  for  the 
adiabatic  as  for  the  isothermal  curve.  This  is  in  accordance  with 
the  physical  ideas  involved  ;  for,  considering  a  compression,  it  is 
clear  that  the  work  done  by  an  external  agent  would  tend  to  raise 
the  temperature  of  the  gas.  If  the  change  is  isothermal,  this  must 
be  prevented  by  conducting  away  the  heat  equivalent  of  the  work 


FUNDAMENTAL   LAWS 


21 


FIG.  3 


as  fast  as  it  is  produced.  The  isothermal  change  is  represented  in 
Figure  3  by  ab.  In  an  adiabatic  compression,  on  the  other  hand, 
this  is  not  done,  and  the  heat  goes 
into  increasing  the  molecular  mo- 
tion of  the  gas.  The  adiabatic  is 
ac  in  Figure  3.  This  increased 
molecular  motion  means  an  in- 
creased pressure  for  the  same  final 
volume  as  was  attained  in  the 
isothermal  compression.  Similar 
reasoning  applies  to  the  case  of  an 
expansion. 

In  general,  it  is  to  be  noted  that 
if  a  compression  is  adiabatic  the 
work  done  goes  into  raising  the 
temperature  and  increasing  the  intrinsic  energy  of  the  gas.  If  an 
expansion  is  adiabatic,  part  of  the  intrinsic  energy  of  the  gas  is 
transformed  into  external  work  and  the  temperature  of  the  gas 
is  lowered.  In  both  cases  the  amount  of  work  done  is  numerically 
equal  to  the  change  in  intrinsic  energy.  On  the  other  hand,  if  the 
change  is  isothermal,  heat  energy  must  be  given  off  during  a  com- 
pression and  absorbed  by  the  gas  during  an  expansion,  and  the 
amount  of  this  heat  energy  in  the  case  of"  a  perfect  gas  is  equivalent 
to  the  work  done,  as  is  stated  in  Joule's  Law.  Thus,  an  isothermal 
transformation  for  most  practical  gas  problems  may  be  considered 
an  isoenergic  transformation. 

23.  Second  Law  of  Thermodynamics.  It  was  stated  in  Section  19 
that  for  a  reversible  'process  an  adiabatic  transformation  was  also  a 
so-called  isentropic  transformation.  The  concept  of  entropy  may 
best  be  developed  after  a  discussion  of  a  cycle  of  changes  known 
as  the  "  Carnot  cycle  "  and  of  a  law  known  as  the  "  Second  Law  of 
Thermodynamics." 

The  Second  Law  of  Thermodynamics  is  similar  to  several  laws  of 
physics  in  that  it  admits  of  110  direct  verification  and  applies  essen- 
tially to  ideal  cases.  The  first  Newtonian  law  of  motion  is  of  this 
kind,  for  it  states  that  "  every  body  continues  in  a  state  of  rest  or  of 
uniform  motion  in  a  straight  line  except  in  so  far  as  compelled  by 
force  to  change  this  state."  The  condition  of  a  body  free  from  the 
action  of  forces,  either  of  friction  or  of  gravitational  attraction,  is 
one  that  is  not  reproducible  in  nature,  and  yet  the  law  is  accepted 
as  essentially  axiomatic,  because  as  nearly  as  the  conditions  can  be 


22 


THERMODYNAMICS 


produced  the  law  is  in  accord  with  known  facts.  The  Second  Law 
of  Thermodynamics  may  be  accepted  in  somewhat  the  same  manner, 
not  because  it  is  capable  of  direct  verification  but  because  no  deduc- 
tion from  it  that  is  capable  of  experimental  proof  has  ever  failed  of 
verification. 

The  Second  Law  of  Thermodynamics  states  that  "  it  is  impossible 
for  any  self-acting  machine,  unaided  by  any  external  agency,  to 
convey  heat  from  a  body  at  a  low  temperature  to  one  at  a  high 
temperature ;  or  heat  cannot  of  itself  (that  is,  without  the  perform- 
ance of  work  by  some  external  agency)  pass  from  a  cold  to  a  warmer 
body."  This  law  applies  only  to  a  type  of  operations  known  as 
" cyclic."  Its  discussion  will  be  postponed  until  after  a  description 
of  the  "  cycle  "  of  changes  suggested  by  Carnot. 

24.  Carnot  Cycle.  Imagine  a  working  substance,  as  for  example 
air  or  gas,  to  be  contained  in  a  cylinder  fitted  with  a  piston.  The 

sides  of  the  cylinder 
and  the  piston  are 
supposed  to  be  abso- 
lute non-conductors  of 
heat.  But  the  bottom 
of  the  cylinder  is  as- 
sumed to  be  a  perfect 
conductor  of  heat. 
Let  there  be  imagined 
also  two  sources  of 
heat  energy,  dia- 
grammed as  A  and  B 
in  Figure  4,  which 
are  maintained  at  con- 
stant temperatures  of 
6l  and  02,  where  0l  is 
higher  than  #2.  Let  there  also  be  provided  an  insulating  stand 
shown  at  C.  Let  the  condition  of  the  gas  in  the -cylinder  be  rep- 
resented by  the  point  a  on  the  pressure-volume  plot  of  Figure  5, 
and  let  this  original  condition  be  so  chosen  that  the  temperature 

Place  the  cylinder,  upon  the  hot  source  A  and  allow  the  gas  to 
expand  isothermally  from  its  initial  state  of  pv  V\,  0V  to  a  state  p^ 
Fg,  #j,  represented  by  the  point  b.  During  this  expansion  .the 
energy  necessary  to  do  the  external  work  against  the  pressure  of  the 
piston  is  derived  from  the  source  A.  Now  remove  the  cylinder 


e, 


FIG.  4 


FUNDAMENTAL   LAWS 


23 


FIG.  5 


from  A  and  placing  it  upon  the  insulating  stand  (7,  allow  an  adiaba- 
tic  expansion  until  the  temperature  has  fallen  to  that  of  the  cold 
source  J5,  namely,  #2.  The  point  c  of  coordinates  pz  and  VB  repre- 
sents on  the  pressure-volume 
plot  the  state  of  the  gas. 
Now  place  the  cylinder  upon 
the  source  B  and  compress 
isothermally  until  the  state 
of  the  gas  is  represented  by 
d  of  coordinates  p±  and  V^ 
and  temperature  #2.  The 
point  d  is  conditioned  by  the 
requirements  that  it  shall 
lie  upon  an  adiabatic  through 
a  as  well  as  upon  the  iso- 
thermal through  c.  Then 
removing  the  cylinder,  place 
it  upon  the  stand  C  and 
compress  the  contents  adia- 
batically  until  the  temperature  has  risen  to  6^  when  the  state  of 
the  gas  will  again  be  represented  by  the  point  a. 

This  set  of  four  successive  operations  is  called  a  cycle  for  the 
obvious  reason  that  the  working  substance  undergoes  a  series  of 
changes  which  return  it  to  its  original  condition.  The  cycle  is 
known  as  the  Carnot  Cycle  from  its  proposer,  and  the  scheme  here 
presented  is  known  as  a  Carnot  Engine..  In  this  ideal  heat  engine 
the  emphasis  is  placed  upon  the  working  substance  and  its  cycle  of 
changes,  and  not  upon  the  mechanical  details  by  which  such  an 
engine  could  be  used  practically. 

These  four  operations  will  now  be  examined  in  more  detail.  Dur- 
ing the  isothermal  expansion  from  a  to  b  an  amount  of  heat  equal  to 
A  x  area  abfe  has  been  received  from  the  hot  source  and  converted 
into  work.  During  the  adiabatic  expansion  from  b  to  c  an  amount 
of  intrinsic  energy  equal  in  heat  units  to  A  X  area  bcgf  has  been  con- 
verted into  mechanical  work.  This  has  been  at  the  expense  of  the 
temperature,  as  noted  above.  During  the  isothermal  compression 
from  c  to  d  an  amount  of  external  work  has  been  converted  into 
A  x  area  cdhg  heat  units  and  abstracted  from  the  gas  by  the  cold 
source  B.  During  the  adiabatic  compression  from  c?  to  a  an  amount 
of  heat  equal  to  A  x  area  daeh  has  been  given  to  the  working  sub- 
stance as  the  result  of  the  external  work,  and  the  temperature  has 


24  THERMODYNAMICS 

consequently  risen.     The  net  result  of  the  cycle  of  operations  has 
been  therefore  the  performance  of  W  units  of  work  by  the  gas,  where 

W=  area  (abfe  -f  bcgf  —  cghd  —  dhea)  =  area  abed. 

Further,  since  the  molecular  energy  possessed  by  the  working 
substance  is  the  same  at  any  point  on  an  isoenergic  and  hence  iso- 
thermal line,  it  follows  that  the  change  produced  in  the  intrinsic 
energy  in  passing  from  one  isothermal  to  another  is  independent  of 
the  path  followed.  Hence,  it  is  considered  evident  that  the  loss  in 
intrinsic  energy  which  occurs  during  the  adiabatic  expansion  from 
b  to  c  is  numerically  equal  to  the  gain  occasioned  by  the  compression 
from  d  to  a.  Representing  by  Qj  the  entire  amount  of  heat  energy 
taken  from  the  hot  source  A  and  equal  to  A  x  area  ab/e,  and  simi- 
larly representing  by  Q2  the  heat  energy  rejected  to  the  cold  source 
B,  of  amount  A  X  area  cdhg,  we  have,  since  the  working  substance 
returns  to  its  original  state,  the  net  amount  of  heat  supplied  equal  to 
the  net  amount  of  external  work  performed.  Then  Qj  —  Q2  =  A  W. 

The  efficiency  of  the  engine  is  to  be  written  as  the  ratio  of  the 
output  A  W  or  Qj  —  £}2,  to  the  input,  which  is  Qr  If  77  represent 
the  efficiency,  we  have  then 

'-if- V1 :.:  .v'-; 

25.  Reversible  Cycle.     The  Carnot  cycle  is  reversible,  inasmuch 
as  it  is  possible  to  perform  these  operations  in  the  reverse  order,  and 
to  return  the  working  substance  to  its  initial  state.     Thus  starting 
with  the  gas  in  condition  represented  by  «,  allow  it  to  expand  adia- 
batically  to  the  state  represented  by  d  in  the  pressure-volume  plot. 
Then  placing  it  upon  the  cold  source  B,  allow  the  gas  to  expand 
isothermally  from  d  to  <?,  while  receiving  an  amount  of  heat  Q2. 
Then  compress  adiabatically  from  c  to  6,  where  the  temperature  is  6^. 
And  finally  placing  the  cylinder  upon  the  hot  source  A,  compress 
isothermally  from  b  to  a  and  reject  to  the  source  an  amount  of  heat 
Oj.     An  amount  of  external  work  TlT,  equal  in  value  to  e/^Qj  —  Q2), 
will  have  been  done  upon  the  gas,  and  an  amount  of  heat  £}2  ab- 
sorbed from  the  cold  source  and  rejected  to  the  hot  source.     This 
transfer  of  heat  from  a  cold  source  to  a  warmer  source  has  been  at 
the  expense  of  work  performed  by  some  external  agency,  and  does 
not  therefore  constitute  a  violation  of  the  Second  Law. 

26.  Carnot's  Theorem.     Upon  the  assumption  of  the  Second  Law 
a  theorem  advanced  by  Carnot  may  be  proved.     This  theorem  states 
that  all  reversible  heat  engines  have  the  same  efficiency  when  absorb- 


FUNDAMENTAL   LAWS  '  25 

ing  and  rejecting  heat  at  the  same  limiting  temperatures.  To  prove 
this  statement  imagine  two  heat  engines,  that  is,  two  working  sub- 
stances, operating  in  the  fashion  described  in  Section  24.  Let  one 
of  these,  denoted  by  A,  have  a  greater  efficiency  than  the  other 
engine,  which  will  be  represented  by  B.  Imagine  the  mechanical 
details  so  arranged  that  when  working  with  the  same  hot  source 
and  the  same  cold  source  the  work  done  by  A  shall  equal  that  done 
by  B.  Then  let  B  pass  through  its  cycle  in  the  reversed  direction, 
in  a  manner  similar  to  that  described  in  Section  25,  and  let  the 
mechanical  energy  necessary  to  drive  it  be  supplied  by  A.  The 
work  done  by  A  will  then  just  equal  that  required  to  drive  B, 
reversed.  Let  A  absorb  from  the  hot  source  an  amount  of  heat  Ox, 
and  reject  to  the  cold  source  an  amount  Q2.  Similarly,  let  B  absorb 
from  the  cold  source  an  amount  Q2',  and  reject  to  the  hot  source  an 
amount  Q/.  Then  if  the  heat  taken  from  a  source  be  indicated  by 
a  minus  sign  and  that  rejected  to  a  source  by  a  plus  sign,  the  net 
result  of  the  two  operations  upon  the  heat  sources  is  shown  in  the 
following  tabular  arrangement  : 

HOT  SOURCE  COLD  SOURCE 

Engine  A  -  Qx  +  Q2 

Engine  B  +  Q/  -  Q2' 

Total  heat  Q^  -  Ox  O2  -  G2' 

Now  since  the  work  done  by  A  is  equal  to  that  which  B  would  do  if 
passing  through  its  cycle  in  the  positive  direction,  it  follows  that 

G!  -  o2  =  Q/—  cy.     .  -v  >•:;•-•   0) 

Also,  since  by  assumption  the  efficiency  of  A  is  greater  than  that 


Combining  equations  (i)  and  (ii)  gives 


and  hence  O/  >  Qr 

Further,  by  comparing  equations  (i)  and  (iii)  it  is  evident  that 

In    the   tabular   arrangement   above    it    appears   therefore    that   an 
amount  of  heat  is  taken  from  the  cold  source  and  added  to  the  hot 


26  THERMODYNAMICS 

source.  The  net  result  is  that  a  self-acting  machine  (or  combina- 
tion of  machines)  performing  a  cycle  is  transferring  heat  from  a  cold 
body  to  a  hot  body  without  the  performance  of  work  by  some  ex- 
ternal agency.  This  is  impossible,  according  to  the  Second  Law. 
Therefore  £)/  cannot  be  greater  than  Qj  and  also  Q2'  cannot  be 
greater  than  &%. 

Hence  the  efficiency  of  engine  A  cannot  be  greater  than  that  of 
engine  B.  A  similar  process  of  reasoning  will  show  that  the  effi- 
ciency of  B  cannot  be  greater  than  that  of  A.  Hence  the  efficiencies 
of  any  two  (or  of  all)  reversible  heat  engines  absorbing  heat  from 
the  same  hot  source  and  rejecting  heat  to  the  same  cold  source  must 
be  the  same. 

27.  Discussion  of  Reversible  Processes.     Planck  says  of  reversible 
processes  i1  "  A  process  which  can  in  no  way  be  completely  reversed 
is  termed  irreversible;  all  other   processes  are  reversible.     That  a 
process   may   be   irreversible   it   is   not  sufficient  that  it  cannot  be 
directly  reversed.     This  is  the  case  with  many  mechanical  processes 
which  are  not  reversible.     The  full  requirement  is  that  it  be  impossi- 
ble, even  with  the  assistance  of  all  agents  in  nature,  to  restore  every- 
where the  exact  initial  state  when  the  process  has  once  taken  place. 
The  generation  of  heat  by  friction,  the  expansion  of  a  gas  without 
the  performance  of  external  work,  the  absorption  of  external  heat, 
and  the  conduction  of   heat,  and  so  on,  are  irreversible  processes. 
Since  there  exists  in  nature  no  process  entirely  free  from  friction  or 
heat  conduction,  all  processes  which  actually  take  place  in  nature  are 
in  reality  irreversible;  reversible  processes  form  only  an  ideal  limit- 
ing case.     They  are,  however,  of  considerable  importance  for  theo- 
retical demonstration  and  for  application  to  states  of  equilibrium." 

28.  Discussion  of  the  Second  Law.     In  a  summary  of  the  founda- 
tions of  thermodynamics  Bryan  says  :2  "  Passage  of  heat  from  one 
body  to  another  is  usually  irreversible  and  therefore  accompanied 
by  a  loss  of  available  energy.     If  we  define  A  to  be  hotter  or  colder 
than  B  according  as  available  energy  is  lost  or  gained  by  the  trans- 
ference of  heat  from  A  to  B,  it  follows  that  heat  can,  and  in  general 
will,  pass  from  hotter  to  colder  bodies,  but  the  reverse  change  can 
only  be  effected  by  combining  it  with  a  compensating  transforma- 
tion.    Carnot's  cycle  reversed   is  a  compensated   reversible   trans- 
formation by  which  heat  can  be  continuously  taken  from  a  colder 
and  given  to  a  hotter  body,  or  vice  versa,  without  loss  of  availability. 

1  Planck,  "  Thermodynamics,"  translated  by  Ogg,  p.  85. 

2  Bryan,  "  Thermodynamics,"  Teubner,  1907. 


FUNDAMENTAL   LAWS  27 

In  this  case  the  compensating  transformation  takes  the  form  of 
work  absorbed  or  produced." 

The  conclusion  of  Carnot's  Theorem  given  in  Section  26  is  often 
stated  as  the  Second  Law.  That  theorem,  however,  holds  only  for 
reversible  cyclic  processes  and  is  not  as  broad  a  statement  of  the 
Second  Law  as  that  quoted  in  Section  23,  which  is  due  chiefly  to 
Clausius.  An  excellent  statement  of  the  essence  of  the  law  is 
quoted  from  Drude1  as  follows:  "Mechanical  work  can  never  be 
continually  obtained  at  the  expense  of  heat  if  only  one  source  of 
heat  at  uniform  temperature  is  at  disposal." 

From  this  statement  there  may  easily  be  derived  an  alternative 
proof  of  Carnot's  Theorem.  Thus  consider  the  engines  A  and  B  of 
Section  26,  in  which  engine  A  is  assumed  to  be  of  greater  efficiency 
than  engine  B,  to  be  so  arranged  mechanically  that  they  would  each 
take  from  the  same  hot  source  the  same  amount  of  heat,  but  because 
of  greater  efficiency  A  would  convert  more  into  work  and  reject  less 
to  the  cold  source  than  would  B.  Now  imagine  engine  A  to  drive 
engine  J5,  reversed,  and  let  B  give  up  to  the  hot  source  just  as 
much  heat  as  A  absorbs  from  it.  Then  the  hot  source  may  be 
omitted  from  consideration  and  B  supply  A  directly.  Because  A, 
has  a  higher  efficiency  it  will  reject  less  heat  to  the  cold  source  than 
will  B  with  its  lower  efficiency  abstract  from  it.  Further,  A  will  do 
more  work  than  will  B,  and  hence  there  will  be  a  net  amount  of 
work  available  for  external  purposes.  But  this  would  be  in  viola- 
tion of  the  Second  Law,  for  this  external  work  would  be  done  as  a 
result  of  a  net  amount  of  heat  absorbed  from  the  cold  source  ;  that  is, 
from  the  single  source  of  heat  at  disposal.  Hence  there  can  be  no 
external  work.  Hence,  further,  the  efficiencies  must  be  the  same. 

In  this  expression  of  the  substance  of  the  Second  Law  by  Drude 
the  emphasis  may  for  the  moment  be  placed  upon  the  word  "  contin- 
ually," which  is  essentially  equivalent  to  the  term  "in  a  cyclic 
process."  As  an  illustration  of  the  fact  that  the  Second  Law  holds 
only  for  cyclic  processes,  consider  the  following  example.  Imagine 
two  non-conducting  cylinders,  one  containing  air  at  a  high  pressure 
and  a  low  temperature  and  the  other  air  at  a  low  pressure  and  a  high 
temperature.  If  they  are  fitted  with  connecting  pistons  so  that  the 
higher  pressure  air  may  expand  and  compress  that  in  the  other  cyl- 

i  Drude,  "  Theory  of  Optics,"  translated  by  Mann  and  Millikan,  pp.  493-494.  This  state- 
ment is  equivalent  to  that  of  Lord  Kelvin  ;  namely,  "  It  is  impossible,  by  means  of  inanimate 
material  agency,  to  derive  mechanical  effect  from  any  portion  of  matter  by  cooling  it  below 
the  temperature  of  the  coldest  of  surrounding  objects." 


28  THERMODYNAMICS 

inder,  then,  for  this  adiabatic  change  the  temperature  in  the  high 
pressure  cylinder  will  fall,  while  that  in  the  low  pressure  cylinder 
will  rise.  During  the  process,  then,  heat  has  been  abstracted  from 
the  low  temperature  side  and  given  to  the  high  temperature  side. 
This  does  not,  however,  constitute  a  violation  of  the  Second  Law, 
since  the  change  is  not  cyclic.1 

Imagine  that  the  low  pressure  cylinder  is  in  connection  with 
the  atmosphere  as  in  a  vacuum  pump.  The  high  pressure  side 
then  does  external  work  in  pushing  out  the  atmosphere  contained  in 
the  pump,  and  in  so  doing  the  working  substance  in  the  high  pres- 
sure cylinder  is  cooled  still  farther  below  the  temperature  of  the 
surrounding  atmosphere.  Heat  may  then  be  absorbed  from  the 
atmosphere  and  the  original  temperature  thus  regained.  But  still 
the  working  substance  is  not  returned  to  its  original  state,  and  to 
return  it  would  require  as  much  work  as  was  originally  obtained 
from  its  expansion.  The  example  cited  is  then  in  accordance  with 
the  Second  Law. 

In  general,  it  may  be  said  for  the  production  of  mechanical  work 
two  sources  of  heat  at  different  temperatures  are  required,  and  the 
production  of  work  requires  the  transfer  of  heat  in  such  a  direction 
as  to  tend  to  result  in  a  final  equality  of  temperature  of  the  sources, 
in  which  condition  all  the  available  energy  of  the  system  will  have 
been  used  up. 

29.  Thermodynamic  Scale  of  Temperature.  For  a  reversible  cycle 
it  has  been  seen  that  the  efficiency  of  all  heat  engines  is  the  same 
provided  only  that  the  temperature  limits  are  the  same.  Lord 
Kelvin  therefore  proposed  a  thermodynamic  scale  of  temperatures 
which  is  independent  of  the  substance  used. 

Draw  any  two  adiabatic  lines  for  a  substance  and  also  two  iso- 
thermal lines  (for  reasons  evident  later  the  isothermals  will  be  those 
for  temperatures  of  0°  and  100°  C.).  Let  the  area  bounded  by  the 
two  isothermals  and  the  two  adiabatics  be  divided  into  100  equal 
parts  by  drawing  other  isothermals.  Calling  the  area  between  any 
two  adjacent  isothermals  a,  divide  the  space  below  the  zero  degree 
Centigrade  isothermal  and  between  the  two  adiabatics  into  areas 
equal  to  a  by  drawing  other  isothermals.  In  Figure  6  this  has  been 
represented  by  drawing  the  isothermals  dividing  the  space  between 
the  0°  and  the  100°  isothermal  into  four  parts  and  extending  this 
division  below  the  0°  isothermal  as  far  as  is  possible  in  a  finite 

1  In  part  this  illustration  is  derived  from  Edser,  "  Heat  for  Advanced  Students,"  p.  340. 


FUNDAMENTAL   LAWS 


29 


diagram.     Let  the  lowest  isothermal  corresponding  to  the  condition 
of  the  substance  when   there   is  no  intrinsic  molecular   energy  be 


FIG.  6 

called  the  absolute  zero  of  temperature  as  in  Section  15.  Number 
the  isothermals  from  this  absolute  zero,  and  let  the  number  of  any 
isothermal  be  represented  by  T. 

Consider  now  a  Carnot  cycle  composed  of  these  two  adiabatics 
and  any  two  isothermals  as  T±  and  T<J,  where  T^  is  higher  than  T2' . 

— 7 — — ,  where  Q/  and  £}2'  are 


. — 
The   efficiency  of   this  cycle  is   — ^— — 

Q 

measured  as  in  Section  20  by  the  areas  under  the  isothermal  lines. 
The  numerator  of  this  expression  is,  however,  equal  to  the  area 
included  by  the  two  isothermals  and  the  two  adiabatics.  If  now  a 
particular  value  of  zero  be  given  to  T%,  and  if  T±  be  put  equal  to  T> 
then  the  expression  for  the  efficiency  of  the  Carnot  cycle  between 


30 


THERMODYNAMICS 


these  limiting  isothermals  becomes 


^-2  =  ^  =  1.  That  is,  Q  =  Zfc. 

This  means  merely  that  in  traversing  any  isothermal  between  the 
two  limiting  adiabatics,  the  input  is  T  x  a  in  work  units.  Hence 
the  efficiency  of  a  Carnot  cycle  between  two  isothermals  T^  and  T2 
is  given  by  the  expression 


a  - 


(24) 


If  the  numbers  which  have  been  assigned  to  the  isothermals  be 
considered  to  be  temperatures,  it  is  evident  that  we  have  a  method 
for  measuring  temperature  which  is  independent  of  the  substance 
chosen.  The  zero  of  this  thermodynamic  scale  coincides  with  the 
perfect  gas  thermometer  zero,  and  the  size  of  a  degree  is  approxi- 
mately the  degree  Centigrade  as  measured  on  a  hydrogen  thermom- 
eter. If  now  it  can  be  shown  that  the  efficiency  of  a  Carnot  cycle 

— -,  where  0l  and 


with  a  perfect  gas  for  a  working  substance  is  — 

02  are  the  absolute  temperatures  on  a  perfect  gas  thermometric  scale, 

it  will  be  evident  that  the  thermodynamic  scale  of  Lord  Kelvin  is 

identical  with  the  perfect  gas  scale. 

30.    Carnot  Cycle  for  a  Perfect  Gas.     It  has  already  been  shown 

that  the  work  done  by  a  per- 
fect gas  during  the  adiabatic 
expansion  from  b  to  c  of  the 
Carnot  cycle  shown  in  Fig- 
ure 5  (reproduced  here  as 
Figure  7)  is  equal  to  that 
done  upon  the  gas  during  the 
adiabatic  compression  from  d 
to  a.  It  remains  only  to  find 
the  values  of  QI  and  Q2  in 
order  to  arrive  at  an  expres- 
sion for  the  efficiency  of  the 
Carnot  cycle  for  a  perfect 
y  gas.  If  the  efficiency  can  be 
expressed  entirely  in  terms 
of  the  absolute  temperatures, 

then  the  comparison  suggested  in  the  preceding  section  can  be  made. 
According  to  equation  (18)  the  area  under  the  isothermal  a  to  5, 

which  is  equal  to  J&v  is  expressible  as  mRB1  loge  (VJV^).     And 


FIG.  7 


FUNDAMENTAL   LAWS  31 

similarly,  the  heat  rejected  in  passing  along  the  isothermal  from  c  to 
d  is  ^=AmR9z  loge(  VC/V^).     The  efficiency  is  therefore 

AmR0.log,  ^- 

_  JUj  —  :U2 {_£ 


During  the  isothermal  changes 

v  and  also  pc  Fc  =  pd  Vd  =  mR6 


,d       i       , 
Hence  ^L*  =  -/  and 

PaVa        #1 


^  for  the  adiabatic  changes  pb  VbK  =  pc  VCK  and 
Substituting  from  these  adiabatic  equations  gives 


9  d  \  i        2        i        0 

therefore  -=^  =  T=£  or  -=?  =  -=?. 

Substitution  of  this  ratio  in  the  expression  for  the  efficiency  gives 

7?  = 


A  comparison  of  equations  (24)  and  (25)  shows  that  a  perfect  gas 
scale  is  identical  with  the  absolute  thermodynamic  scale. 

31.  Entropy,  In  equation  (23),  for  the  efficiency  of  a  Carnot  per- 
fect gas  cycle,  if  the  two  isothermals  are  taken  at  an  infinitesimal 
difference  of  temperature  apart,  so  that  01  —  02  =  dO,  then  the  corre- 
sponding heat  Oj  —  Qg,  converted  into  work,  may  be  represented  by 
the  infinitesimal  c?Q,  and  by  equation  (25)  the  expression  for  the 
efficiency  becomes 

d&  =  dd 
O  ==  6  ' 


32  THERMODYNAMICS 

The  solution  of  this  differential  equation  gives 

loge^l  =  logA  or  ^1  =  ^1, 
36  02         3e#2'        02      02 

and  hence  §1=?2-      '*•   *     •     '     '     •     •     (26) 

01       02 

This  means  that  for  a  reversible  process  the  ratio  of  the  heat  ab- 
sorbed (or  rejected)  to  the  absolute  temperature  of  the  isothermal 
traversed  in  passing  between  two  adiabatics  is  independent  of  the 
isothermal  traversed.  The  quantity  which  remains  constant  under 
these  conditions  is  the  difference  in  entropy  of  the  substance  in  the 
states  represented  by  the  two  adiabatics.  An  adiabatic  line  is  then 
for  a  reversible  process  by  definition  a  line  of  constant  entropy,  or  an 
isentropic. 

The  entropy  of  any  state  will  be  represented  by  the  usual  symbol 

c/>.     The  expression  —  is  then  an  expression  for  the  difference  in 
6 

entropy,  to  be  denoted  by  $2  —  <£r  between  the  entropy  0X  of  the 
first  adiabatic  and  $2,  the  entropy  corresponding  to  the  second  adia- 
batic. This  change  in  entropy  between  these  two  isentropics  is 
independent  of  the  path  followed,  provided  only  the  path  is  a  rever- 
sible one,  as  it  was  in  the  'case  of  the  isothermal  of  the  Carnot  cycle 
from  which  this  concept  was  developed. 

In  general,  if  a  small  amount  of  heat  c?Q  be  added  to  a  body  at  a 
temperature  0,  which  is  sensibly  constant  during  that  addition,  then 
the  entropy  of  the  body  has  been  increased  by  an  amount  <#<£,  where 


. 

If  the  change  in  entropy  is  produced  by  a  large  number  of  such 
additions,  by  processes  which  are  completely  reversible,  the  total 
change  in  entropy  thereby  produced  is 


(27) 


For  the  special  case  where  the  heat  is  added  along  an  isothermal 
this  reduces  to 


FUNDAMENTAL   LAWS  83 

Further,  it  is  to  be  noticed  that  for  the  Carnot  cycle 

^1-^=0. 

*1    02 

That  is,  for  a  reversible  cycle 

.  (J)^  =  0,        .     .     .    V.     .     (28) 

where  the  symbol  ( J  )  indicates  the  fact  that  the  integration  is  to  be 

performed  about  a  complete  cycle.  This  equation  is  sometimes 
stated  as  a  mathematical  form  of  the  Second  Law,  but  it  is  prefer- 
able to  consider  it  a  deduction  from  it. 

32.  Measurement  of  Entropy.     It  is  not  possible  to  determine  the 

total  entropy  of  a  body.     That  is,  the  expression  J  —^  when  used 

to  find  the  entropy  of  a  body  in  a  given  state  will  always  contain  a 
constant  of  integration.  The  integral  does,  however,  admit  of  the 
determination  of  the  difference  in  entropy  between  two  given  states. 
Hence  it  is  that  entropy  is  always  expressed  as  so  much  more  or  less 
than  the  entropy  of  some  chosen  state  of  the  substance  under  consid- 
eration. This  is  satisfactory  for  all  practical  uses  of  the  concept  of 
entropy.  Having  arbitrarily  selected  some  adiabatic  as  a  reference 
state  to  which  all  other  states  of  the  substance  are  to  be  compared, 
it  is  possible  to  conceive  of  the  entire  pressure-volume  plot,  for  a 
given  mass  of  the  substance,  as  filled  with  adiabatic  or  isentropic 
lines  which  represent  states  of  the  substance  differing  by  units  of 
entropy,  just  as  the  states  represented  by  the  isothermal  lines  differ 
by  units  of  temperature. 

Quoting  from  Preston,1  "The  entropy  of  a  body  being  arbitrarily 
taken  as  zero  in  some  standard  condition  A  defined  by  some  standard 
temperature  and  pressure  (or  volume),  the  entropy  in  any  other  state 

/V7O) 

B  is  the  value  of  the  integral  J  - —  taken  along  any  reversible  path 

by  which  the  body  may  be  brought  to  state  B  from  standard  state  A. 
The  path  may  obviously  be  an  arc  A  0  of  an  isothermal  passing 
through  the  point  J.,  defining  the  standard  state,  together  with  the 
arc  B 0  of  the  adiabatic  line  passing  through  the  point  B." 

33.  Temperature-Entropy  Plot.     It  is  evident  from  equation  (27) 
that,  provided  a  change  is  taking  place  along  a  reversible  path,  the 

i  Preston,  "  Heat,"  p.  628. 


34 


THERMODYNAMICS 


heat  added  (or  rejected)  may  be  expressed  as  the  product  of  a  tem- 
perature term  and  a  difference  in  entropy  term,  thus 


e, 


e 


FIG.  8 


For  this  reason  and  because  of  the  practical  interest  in  steam  en- 
gineering of  following  the  changes  in  entropy  of  steam  it  is  most 

convenient  to  plot  values  of  entropy  on  a 
temperature-entropy  plot. 

In  the  temperature-entropy,  or  "0-0 
plot,"  the  ordinates  are  temperatures  and 
the  abscissas  are  entropies  measured  from 
some  standard  state  as  described  above. 
The  0-(f)  diagram  of  the  Carnot  cycle  will 
now  be  plotted  as  an  illustration.  During 
the  isothermal  expansion  from  a  to  b  of 
Figure  7,  the  entropy  increases  by  an 
amount  by  </>2  —  ^,  and  the  change  is  rep- 
resented as  ab  in  the  plot  of  Figure  8. 
The  reversible  adiabatic  expansion  from  b 
to  c  is  isentropic  by  definition  and  is  rep- 
resented on  the  6-<t>  plot  by  the  line  be. 
The  isothermal  or  constant  temperature 
change  cd  and  the  reversible  adiabatic  compression  da  complete  the 
cycle. 

It  is  to  be  noticed  that  for  this  case,  since  the  isothermal  is  a  re- 
versible process,  the  heat  absorbed  at  temperature  0l  is  represented 
by  the  area  under  the  line  ab.  Similarly,  the  heat  rejected  during 
the  isothermal  compression  from  c  to  d  is  given  by  the  area  under 
the  line  cd.  The  net  heat  converted  into  work  in  this  cycle  is  then 

Q!  —  Q2  =  ^i  ($2  ~~  $1)  ~~  ^2  ($2  ~  $1)  =  (^1 ~~  ^2)  ($2  ~~  0i) '  that  is,  it 
is  represented  by  the  area  enclosed  by  the  0-(f>  plot  of  the  cycle. 

It  is  important  to  note  carefully  that  it  is  only  when  the  trans- 
formation takes  place  in  a  reversible  manner  that  the  heat  absorbed 
(or  rejected)  is  represented  by  the  area  underneath  the  curve  on  the 
0-<f>  diagram  representing  the  transformation.  In  all  other  cases 
the  increase  in  entropy  will  be  disproportionate  to  the  increase  in 
heat  contents  as  in  the  case  of  the  porous  plug  experiment  to  be  de- 
scribed in  Section  35.  Although  this  restriction  limits  this  use  of  a 
0-(f)  plot  to  the  ideal  cases,  such  a  plot  is  still  of  great  interest  and 
value  in  many  practical  problems. 


FUNDAMENTAL  LAWS 


35 


e 


FIG.  9 


34.  Discussion  of  Carnot  Cycle  from  0-<|>  Plot.      If  any  other  re- 
versible cycle  be  plotted  upon  the  #-</>  diagram  as  in  Figure  9,  it 
is  evident  that  the  work  output,  which  would  be  proportional  to  the 
area  enclosed  by  the  plot,  is  less  than  for  the  Carnot  cycle,  although 
the  same  range  of  temperatures  is  used.     It 

follows  then  that  no  reversible  heat  engine 
can  have  an  efficiency  greater  than  that  of  the 
Carnot  engine.     And  of  course  all  heat  en-  ~ 
gines  with  irreversible  transformations  are 
less  efficient  for  the  same  limiting  tempera- 
tures than  the  reversible  engines.     An  ex- 
ample of  a  reversible  cycle  equal  in  efficiency 
to  the  Carnot  cycle  is  given  in  Problem  11,  "z 
on  page  51. 

It  is  evident  from  Figures  8  and  9  that 
the  high  efficiency  of  the  Carnot  cycle  is 
obtained  as  the  result  of  two  conditions, 
namely,  that  all  the  heat  absorbed  is  taken 
in  at  the  highest  temperature,  and  all  the 
heat  rejected  is  at  the  lowest  temperature. 
In  so  far  as  any  heat  engine  conforms  to 
these  requirements  (neglecting  the  fact  that  its  other  transforma- 
tions will  not  in  general  be  isentropic)  will  its  efficiency  approach 
the  limiting  and  maximum  efficiency  of  a  Carnot  engine  working 
between  the  same  temperatures. 

35.  Entropy  Changes  in  Irreversible  Processes.     Returning  again 
to  the  concept  of  entropy,  it  may  be  said  that  entropy  is  that  prop- 
erty of  a  body  which  remains  constant  so  long  as  the  body  is  under- 
going adiabatic  changes  that  are  reversible.     If  the  change  is  not 
reversible,  then  the  entropy  in  the  final   state  is   greater   than  it 
would  be  for  a  reversible  transformation.     This  is  indicated  by  the 
classical  "  porous  plug  experiment "  of  Kelvin  and  Joule.     In  this  ex- 
periment, which  offers  a  more  accurate  method  than  does  that  de- 
scribed in  Section  11  for  testing  Joule's  Law,  it  was  found  that  when 
air  expanded  from  a  higher  pressure  to  a  lower  pressure  without 
doing  external  work  there  was  a  distinct  lowering  of  the  temperature, 
due  to  the  partial  conversion  of  the  kinetic  energy  of  the  molecules 
into  potential  energy  of  molecular  separation.     The  experiment  was 
performed  by  allowing  air  to  expand  from  a  high  pressure  through 
a  plug  of  cotton  wool  to  atmospheric  pressure.     The  plug  was  sur- 
rounded with  a  non-conductor  and  the  expansion  was  adiabatic. 


36 


THERMODYNAMICS 


For  the  purposes  of  the  present  discussion  the  important  point  of 
the  experiment  is  as  follows.  A  mass  of  the  gas  reaching  an  open- 
ing in  the  plug,  upon  the  other  side  of  which  the  pressure  is  lower, 
expands  adiabatically  (a  reversible  and  hence  isentropic  change), 
converting  a  portion  of  its  intrinsic  energy  into  mechanical  kinetic 
energy  of  the  entire  mass.  This  mechanical  kinetic  energy  is  im- 
mediately converted  into  heat  by  friction,  viscosity,  and  the  like. 
The  quantity  of  heat  corresponding  to  this  friction  work  is  now 
returned  to  the  air  and  causes  an  increase  in  entropy,  exactly  as  if 
it  had  been  added  from  some  outside  source.  As  a  result  of  these 
two  infinitesimal  changes,  the  first  isentropic  and  the  second  of 
increasing  entropy,  the  final  entropy  of  the  air  is  obviously  in- 
creased. The  same  conclusions  as  to  an  increase  in  entropy  may 
be  made  for  the  original  experiment  of  Joule,  but  it  is  easier  to  con- 
ceive of  the  steps  in  this  case.  This  expansion,  although  adiabatic, 

is  obviously  irreversible  and 
the  entropy  is  increased.  For 
further  discussion  it  will  be 
assumed  to  have  been  shown 
that  all  irreversible  transfor- 
mations lead  to  a  final  state 
of  greater  entropy  than  would 
a  corresponding  reversible 
transformation. 

Consider  a  change  of  state 
for   a   given   substance   from 
A  to  B  by  a  reversible  path 
as  represented  on  the  pressure- 
V  volume  plot  of  Figure  10  by 

the  line  ArB.      This  change 

might  be  represented  on  the 
temperature-entropy  plot  of  Figure  11  by  the  line  ArB.  If  now 
the  transformation  from  A  to  B  is  along  some  irreversible  path  as 
AiB  in  Figure  10,  it  may  be  represented  on  the  (f>-0  plot  by  some 
line  AiB' ,  since  the  final  entropy  is  greater  than  for  the  reversible 

change.     The  integral     \    — —  gives  the  change  in   entropy  along 

the  reversible  path,  but  is  less  than  the  true  change  along  the  irre- 
versible path.     Thus 

XB  ^7O 
_.  ......        (i) 


FIG.  10 


FUNDAMENTAL   LAWS 


37 


Similarly,  if  the  transformation  take  place  from  B  to  A,  the  entropy 

along  the  reversible  path  is  given  by  the  integral   |    — ,  but  the 

J  B    9 

entropy  in  the  final  state  for  the  irreversible  path  is  greater  than 
that  for  the  reversible  path.     That  is, 


For  the  irreversible  change,  represented  on  the  p-v  plot  by  ArBiA 
the  total  change  in  entropy  is  not  zero  as  it  would  be  for  an  en- 
tirely reversible  process,  but 
is  obviously  greater  than  zero  9 
(or  at  the  most  equal  to  it). 
For,  from  A  to  B  by  a  re- 
versible process 


6  ' 

and  from  B  to  A  by  an  irre- 
versible process 


Then  by  the  path  ArBiA 


4> 


(v) 

*s  V 

That   is,  for   the    irreversible 
process    just    considered,    the 

expression    I  -~  taken  for  the  entire  change  is  less  than  the  actual 
•^    6 

increase  of  entropy.  But  the  physical  state  represented  by  <t>A>  is 
not  the  same  as  the  initial  physical  state  of  entropy  </>A.  In  order, 
then,  that  the  process  shall  be  a  true  cycle,  it  is  necessary  to  sub- 
tract an  amount  of  entropy  of  value  in  this  case  <f)A'  —  <f>A.  This 
fact,  usually  stated  in  the  form  given  below,  is  due  to  Clausius. 

|0 (30) 


CHAPTER   II 

GASES 

36.    Graphical  Representation  of  State  for  a  Perfect  Gas.     It  is  a 

simple  fact  of  solid  analytical  geometry  that  a  function  of  three 
variables  in  general  represents  a  surface.  The  general  equation  for 
a  given  mass  m  of  a  perfect  gas,  namely  pV—  mRO,  is  evidently  a 
function  of  three  variables  and  may  be  taken  as  representing  a  sur- 
face plotted  with  reference  to  three  rectangular  axes,  p,  V,  and  0, 
corresponding  to  a?,  y,  and  2.  This  surface  is  known  as  the  "  char- 
acteristic surface  "  of  the  gas.  Since  the  position  of  any  point  on 
this  surface  represents  a  corresponding  state  or  condition  of  the  gas, 
such  a  point  is  usually  termed  a  "state  point"  of  the  gas.  Now, 
whatever  changes  the  given  mass  of  gas  may  undergo,  its  successive 
states  must  be  represented  by  state  points  on  this  characteristic  sur- 
face, corresponding  to  the  successive  values  of  j9,  F",  and  6  which  the 
gas  assumes  during  its  transformation.  The  continuous  curve  pass- 
ing through  these  points  is  a  curve  in  space  and  has  an  equation 
corresponding  to  the  type  of  transformation  which  the  gas  has 
undergone. 

The  equation  in  rectangular  coordinates  of  a  curve  in  space  is 
usually  expressed  by  two  equations,  each  containing  two  vari- 
ables only.  These  two  equations  represent  two  cylinders,  which 
are  right  projecting  cylinders  of  the  curve  upon  the  two  coordinate 
planes  corresponding  to  the  two  sets  of  variables  in  which  the  equa- 
tions are  expressed.  The  intersection  of  these  two  cylinders  deter- 
mines completely  the  curve  in  space.  In  the  case  of  a  curve  in  space 
representing  a  gas  transformation,  since  the  curve  must  lie  upon  a 
known  surface,  one  equation  in  addition  to  the  equation  for  this 
characteristic  surface  is  sufficient.  This  equation  might  be  expressed 
in  terms  of  p  and  F",  p  and  0,  or  V  and  6. 

The  projection  on  the  p  and  F' plane  is  most  serviceable  for  ordi- 
nary purposes,  as  is  evident  from  the  discussion  of  Sections  19  and 
20.  It  is  to  be  noticed  that  the  expressions  there  given  in  p  and  V 
coordinates  for  various  type  transformations  represent  equations  for 
a  right  projecting  cylinder  of  the  space  curve,  perpendicular  to  the 

38 


GASES  39 

pressure-volume  plane,  and  also  that  the  third  coordinate  has  already 
been  found  from  the  characteristic  equation  by  substitution  of  the 
values  of  p  and  V. 

In  general,  a  characteristic  surface  of  unit  mass  of  any  substance 
may  be  represented  by  the  equation  stating  the  relation  of  any  three 
of  the  five  following  magnitudes :  specific  pressure,  specific  volume, 
absolute  temperature,  intrinsic  energy,  and  entropy.  This  is  true 
because  in  general  if  any  two  of  these  magnitudes  are  known  for  a 
given  state  of  the  substance,  the  values  of  the  remaining  three  for 
this  state  may  be  expressed  as  functions  of  the  first  two.  Two  ex- 
ceptions occur,  one  evidently  in  the  case  of  perfect  gases  (and  for 
practical  purposes  in  the  case  of  all  gases)  inasmuch  as  the  abso- 
lute temperature  of  a  gas  is  directly  proportional  to  its  intrinsic 
energy.  (The  other  exception  is  in  the  case  of  a  vapor  where  the 
temperature  and  the  pressure  do  not  completely  determine  the  other 
magnitudes.) 

The  expression  of  a  transformation  in  terms  of  the  absolute  tem- 
perature and  the  entropy  is  frequently  of  value,  and  the  temperature- 
entropy  plot  will  be  used  throughout  the  remainder  of  this  text.1 
Determinations  of  entropy  and  intrinsic  energy  from  direct  measure- 
ments are  of  course  impossible.  The  three  magnitudes  that  may  be 
directly  measured,  namely,  the  temperature,  pressure,  and  volume, 
are  all  easily  found.  For  this  reason,  as  well  as  because  of  the  rela- 
tion between  work  and  area  on  a  p-v  plot  (see  Section  19),  the 
pressure-volume  plot  is  used  almost  exclusively  in  practical  work. 

37.  Equations  for  Polytropic  Transformation  of  a  Perfect  Gas.  The 
equation  for  the  most  general  type  of  gas  transformation  has  been 
given  in  Section  19  as 

p^V-f  —  p^V^  =  a  constant.    >     .    -.     .     .     (i) 

For  n  equal  to  unity  this  equation  expresses  the  relations  of  an  iso- 
thermal change,  and  for  n  equal  to  K  the  relations  of  an  adiabatic 
change  v 

Since  it  is  frequently  desirable  to  express  some  special  form  of 
such  a  polytropic  transformation  in  terms  of  p  and  #,  or  of  J^and  0, 
general  equations  f*or  that  purpose  will  now  be  derived.  Combin- 
ing equation  (i)  with  the  relation  for  a  perfect  gas,  namely, 


1  A  pressure-entropy  plot  is  sometimes  used  in  superheated  steam  problems,  as  is  dis- 
cussed on  page  96. 


40  THERMODYNAMICS 

gives 

^-1  =  0,^-1, '  .     (iii) 

or 


»-i 

•"        •  (31) 


Also  from  (i)  and  (ii) 

^^? 

hence 


or 


(32) 
ft 

38.  Fundamental  Heat  Equations  of  Perfect  Gas.  Since  a  gas 
transformation  may  be  represented  by  the  projection  of  the  state 
curve  upon  the  plane  of  p  and  F",  p  and  0,  or  V  and  #,  it  follows 
that  the  effect  of  adding  to  a  given  mass  (as  unit  mass)  a  small 
amount  of  heat  d<\  may  be  expressed  in  terms  of  any  two  of  these 
variables.  It  is  to  be  remembered  that  this  has  already  been  done 
in  terms  of  p  and  v  in  Section  21,  where  the  result  was  expressed  in 
equation  (21),  which  is  here  reproduced. 


,  •    •    (21) 

This  equation  will,  however,  be  put  in  its  more  usual  form  by  sub- 
stituting from  the  equation  for  unit  mass  of  a  perfect  gas,  namely, 


pv  = 


the   value   of  -  for  %-  and   also  -  for  —  ,  giving    equation    (33)    as 
v         R  p          II 

below, 


.    ...:  .     .     .     .     (33) 
v  p 


From  equation  (i)  we  have  also 

dv  =  d^ade  +  ^dp  =  ^ 

dt)  dp   *         p 

and 

dp  =  ^dd  +  ^dv  = 

00  dv  v 


GASES  41 

Substituting  from  (ii)  in  (33)  gives  equation  (34),  which  expresses 
in  terms  of  the  temperature,  9,  and  pressure,  p,  the  same  effect  as 
does  (33).  Thus 


R6j~}t       6j  RQ  j*     (    RO*        0\i 

--  -dp    -f  cv-dp  =  cp  —  dO-  (c  p  —  --  cv-)dp. 

2}       J         P  pv          V  Pp2v        vj  1 

(34) 


v_p  2}  P  pv          V    pv         p 

Hence 


Similarly,  substituting  from  (iii)  in  (33)  gives  dq  in  terms  of  v  and  0. 
Thus 


(35) 


39.  Entropy  Change  for  Perfect  Gas.  From  the  three  equations 
derived  above  there  are  easily  obtained  three  expressions  for  the 
difference  of  entropy  between  two  states  of  a  gas  provided  the  change 
produced  may  be  considered  reversible.  In  case  of  a  reversible 
transformation 


If  the  coordinates  of  state  1  be  represented  by  pv  v1  ,  and  0V  and 
similarly  the  coordinates  of  state  2  by  p%,  #2,  and  #2,  then  we  may 
obtain  from  equation  (33)  the  following  expression  for  the  differ- 
ence in  entropy: 


dv~\    cp*  dp 

—     =    I       Cv— 

V  J       »/Pf          p 


or 


^  -^=cv  log  ea  +e  p  loge^   .     ...     .     (36) 
Pi  vi 

And  similarly,  equations  (34)  and  (35)  lead  to  equations  (37)  and 
(38),  as  follows  : 

&  -  *i  =  e,  loge     +  (cp  -  cv)  loge^          ,.     i     (3T) 


.    .     .'   V    (38) 


Any  one  of  these  three  equations  may  be  used  to  calculate  the 
change  in  entropy.  Thus  if  the  change  is  isothermal,  (37)  and  (38) 
both  reduce  to  a  simple  form,  since  the  logarithm  of  the  term 
involving  the  temperature  reduces  to  zero  in  this  case.  Similarly, 


42 


THERMODYNAMICS 


for  a  constant  volume  change,  (36)  or  (37)  may  most  easily  be  used. 
It  is  important  to  notice  that,  since  these  equations  all  involve  the 
specific  heat  and  were  derived  for  unit  mass  of  the  gas,  the  values  of 
the  entropy  found  by  their  use  is  in  every  case  per  unit  mass  of  the 
substance,  and  to  obtain  the  total  entropy  change  the  value  <f>2  —  ^ 
must  be  multiplied  by  m.  With  this  understanding  it  makes  no 
difference  whether  the  ratio  of  the  volumes  is  v^v2  or  V<JVV  since 
these  are  equivalent  ratios. 

40.    Relation  of  the  Specific  Heats  of  a  Gas.     It  has  already  been 
stated  in  Section  18  that  cp/cv  is  a  ratio,  /e,  that  depends  upon  the 

nature  of  the  gas.  It 
will  now  be  shown  that 
CP  —  cv  =  AR,  where  R  is 
the  constant  of  the  char- 
acteristic gas  relation. 

If  unit  mass  of  a  per- 
fect gas  at  a  pressure, 
p,  and  a  volume,  v,  is 
heated  from  6l  to  02 


P, 


-v—  - 

\e,     1 
\ 

\ 
\ 
\ 
\ 


\ 


\e2 


\ 


\ 


\ 


V, 

FIG.  12 


under  the  conditions  of 
constant  volume,  the  p-v 
plot  of  the  change  will 
be  as  shown  in  Figure  12 
by  the  line  ac.  The  heat 

'    supplied    is    cv(Q%  —  0^). 

Since  no  external  work 
is  done  by  the  gas,  this 
expression  represents  in  heat  units  the  increase  in  intrinsic  energy 
of  the  gas.  If,  however,  the  change  from  6l  to  02  be  made  at  con- 
stant pressure,  as  represented  by  the  line  ab  in  the  figure,  then  the 
heat  added  is  cp(62  —  0^).  Since  the  final  temperatures  are  the  same, 
the  increases  in  intrinsic  energy  are  the  same.  During  the  heating 
at  constant  pressure  an  amount  of  external  work  represented  by  the 
shaded  area  of  the  figure  has  been  done.  The  value  in  thermal  units 
of  this  external  work  is  Ap(v2  —  v^).  It  is  evident,  then,  that 


-  0  ) 


-  #1)  =  Ap^  -  »,), 


(0 


or 


but 


*,-*, 

pv^  =  R9l  and 


.     .     .     .    (iii) 


GASES 


43 


From  (iii)  we  have 


Hence  substituting  from  (iv)  in  (ii)  gives 


P, 


(39) 

41.  Changes  in  Intrinsic  Energy  of  a  Gas.  For  a  perfect  gas  the 
intrinsic  energy  consists  entirely,  as  has  been  said  in  Section  15,  of 
molecular  kinetic  energy  and  is 
directly  proportional  to  the  abso- 
lute temperature.  The  value  of 
the  intrinsic  energy  of  a  gas  at 
a  given  pressure  and  occupying  a 
given  volume  may  then  be  ex- 
pressed in  terms  of  this  pressure 
and  volume  in  accordance  with 
the  following  reasoning.  Imagine 
the  gas  to  expand  adiabatically 
from  the  state  a,  at  which  the 
value  of  the  intrinsic  energy  is 
desired,  until  the  pressure  is  zero. 
During  this  expansion  all  the  in- 
trinsic energy  is  converted  into 
external  work.  Such  an  expansion  is  represented  diagrammatically 
in  Figure  13.  The  energy  may  therefore  be  measured  by  the  area 
under  the  adiabatic  drawn  through  a  as  in  the  figure.  Denoting 
the  energy  in  state  a  as  E-^  we  have,  according  to  equation  (19)  of 
Section  20,  f 

In  this  equation  the  volume  V^  corresponding  to  zero  pressure',  will 
be  infinite  and  hence  the  expression  for  E±  reduces  to 


FIG.  13 


This  may  also  be  written  as 

*~  7? 

•ffi  = 


K-l 


(40) 


(41) 


From  this  form  it  is  evident  that  the  value  given  in  equation  (40) 
contains  the  mass  implicitly,  and  hence  the  above  expressions  for 
intrinsic  energy  are  for  the  total  energy  and  not  per  unit  mass  of  the 
gas. 


44  THERMODYNAMICS 

The  difference  in  intrinsic  energy  between  two  states  1  and  2  of 
the  same  mass  of  gas  is  evidently 


K  —  1 

In  order  that  this  energy  shall  be  expressed,  in  foot  pounds  it  is  neces- 
sary that  p  be  expressed  in  pounds  per  square  foot  and  V  in  cubic  feet. 

42.  Imperfect  Gases.     As  has  been  previously  stated,  for  ordinaiy 
engineering  purposes  it  is  sufficient  to  apply  the  relations  established 
in  this   chapter   for   perfect  gases  to  any  of  the  ordinary  so-called 
"  permanent  gases  "  such  as  hydrogen,  oxygen,  nitrogen,  marsh  gas, 
and  mixtures  like  air  (and  also,  to  some  extent,  to  carbon  dioxide 
and  the  like).     When  it  is  desired  to  analyze  the  action  of  any  mix- 
ture, it  may  frequently  be  done  by  the  use  of  Dalton's  Law  of  partial 
pressures,  which   states   that  if  pl  and  p%  represent  the  pressures 
which  two  chemically  inert  gases  would  exert  at  the  same  tempera- 
ture when  separately  occupying  a  volume  V,  then  the  resultant  press- 
ure p  for  that  'temperature  due  to  both  these  gases  confined  in  this 
volume  is  the  sum  of  pl  and  p2. 

Certain  constants  for  the  ordinary  gases  are  to  be  found  in  Table  2 
on  page  130. 

43.  Resume  of  Equations  for  Cases. 

The  numbers  following  the  equations  are  the  original   numbers 
assigned  to  them  in  the  preceding  chapters. 
Temperature. 

6  =  459.5  +  t  Fahrenheit  ;  0  =  273  +  t  Centigrade.    .     .     (11,  12) 
External  Work. 


Characteristic  Equation. 

pV=mR6  .....     .     .     .     (13) 

Specific  Heat.      , 

<*=  e 

£fit=  *.....;       (Section  18) 
«• 
cp-cv=AB.  ......      .     ,     (39) 

G-eneral  Heat  Equatidns. 

(1) 

(29) 


GASES  45 

W)=A(dI!+dW).     .     .  (4) 

°  +  cj9&  .......  (33) 

v  p 

(cp-cv)6^.     .....  (35) 

dn  =  Cfae-(cp-Cv)e&.       .,..:"..-;:.  (34) 

Isothermal  Transformation. 

P^ri=P^  ........  (5) 

W=pl^loge^=piri\oge^==mlieiloge^      -  (18) 
v\                      Pz                         '  i 

Adiabatic  Transformation. 

n    T/  K  _  nr.    77"  K  ^99^ 

Pi  '  i  —  Pi  yz  -       •     •     •  ,  •     •     •  \^) 


K:  — 


Intrinsic  Energy. 

•     •  -•    ,     •     •     (40) 


.          .         >      ;-.          ..-          (27) 

V 

A+^loge-^2.       -     •     •     (36) 
'Pi  '   ri 

_  0^         -  \  1       .        *&  fQ7\ 

i  t/o          s  "\  1  IP*)  /^QQ\ 

Carnot  Cycle. 

-         Q1^Q2  =  ^2.      ....    (23,25) 


46  THERMODYNAMICS 


PROBLEMS   AND   SOLUTIONS:    GASES 

1.  Use  of  the  characteristic  equation.      A   gas   tank  COD  tains  1.5 
pounds  of  oxygen  at  a  temperature  of  65°  Fahrenheit  and  a  gauge 
pressure  of  50.3  pounds  per  square  inch.     The  atmospheric  pressure 
is  standard  atmospheric  pressure.     Find  how  much  more  oxygen  the 
tank  can  contain  and  not  exceed  its  safe  pressure  limit  of  250  pounds 
gauge,  if  the  temperature  is  liable  to  rise  to  100°. 

Solution.  A  gauge  always  reads  the  difference  between  the  pressure  which  it  is  to 
measure  and  the  atmospheric  pressure.  Normal  or  standard  atmospheric  condition  is 
a  pressure  of  14.7  pounds  per  square  inch. 

First,  find  the  volume  of  the  tank  from  eq.  (13)  and  Table  2, 

p=  (50.3+  14. 7)  (144),          0  =  459.5  +  65  =  524.5,          m  =  1.5,        72  =  52.56. 

.    y=  (1.5)(52.56)(524.5)  = 
(65) (144) 

Then,  find  the  weight  of  oxygen  which  the  tank  can  contain  under  the  conditions 
p  =  (250  +  14.7)  (144),          e  =  459.5  +  100  =  559.5, 

w  =  (264.7)(144)(4_I42)  =  • 
(52. 56)  (559. 5) 

The  added  weight  is  then  5.73  -  1.6  =  4.23  pounds. 

2.  Oarnot  cycle.     A  Carnot  engine  working  between  350°  Fahren- 
heit and  75°  does  116,700  foot  pounds  of  work  per  cycle.     Find  how 
much  heat  is  supplied  and  rejected  per  cycle. 

Solution.          0i  =  459.5  +  350  =  809.5,        02  =  459.5  +  75  =  534.5, 
el  —  02     809.5  —  534.5     n  q, 
-07"  809.5 

W  =  116.700  foot  pounds,        AW  =  ^  (116,700)  =  150  B.  t.  u. 

•nBi  =  AW.  .-.  Oi  =^||  =  442  B.  t.u. 

Qi-Q2  =  AW.  .'.  02  =  442  -  150  =  292  B.  t.  u. 

3.  Intrinsic  energy.     What  is  the  energy  of  2  pounds  of  air  at 
150  pounds  per  square  inch  pressure,  absolute,  if  it  occupies  6  cubic 
feet  ? 

Solution.         p  =  (150)  (144),          F=6,        m  =  2,          K=  1.40. 
By  eq.  (40),         E  =  ^-  •     .-.  E=  (150)(144)(6)  =  324,000  foot  pounds. 

/C  ~~~  1  J..4U  —  J. 

Alternative  solution.  Find  the  temperature  0,  and  assuming  that  the  value  of  cv 
holds  from  the  absolute  zero  to  this  temperature,  calculate  the  number  of  thermal  units 
to  raise  the  temperature  of  2  pounds  of  air  from  zero  to  0.  Then  reduce  thermal  units 
to  mechanical  energy  units. 


GASES  47 


E  =  mcve  =  (2)  (0.  169)  (1215)  =  411  B.  t.  u. 
.-.  E  =  (411)  (778)  =  320,000  foot  pounds.1 

4.  Constant  pressure  transformation.  If  2  pounds  of  air  are 
heated  at  constant  pressure  from  60°  Fahrenheit  to  200°  find  how 
much  external  work  is  done  and  how  much  the  intrinsic  energy  is 
increased.  How  much  heat  is  required  ? 

Solution.  Cp  =  0.2375,  cp  —  cv  =  AB,  R  =  53.35,  m  =  2,  t\  =  60,  t-2  =  200. 
The  heat  required  at  constant  pressure  is  cpm(t2—  h)=  (0.2375)  (2)  (200—  60)  =  66.5  B.  t.u. 


External  work  done is(cp-cv)(m)(«2-«i)=^2?(2)(200-60)=  ^^ 

778  

Increase  in  intrinsic  energy  is      .        .        ."       .        .         .        .  .     47.3 B.  t.u. 

Or,  expressed  in  mechanical  units,  it  is  (47.3)  (778)  =36,800  ft.  Ib. 

5.  Exponent  n.     Given  2  pounds  of   air  at  a  pressure  of  14.7 
pounds  absolute  which  occupy  a  volume  of  24.78  cubic  feet.     If  the 
air  expands  until  the  pressure  in  0.4^  and  the  volume  is  F^  =  2  Fj, 
find  the  exponent  n  in  the  equation  pVn=plV{1,  which  may  be 
taken  to  represent  the  expansion. 

Solution.    piVin  =  p2  F2n,  where  p2  =  OApi  and  F2  =  2  FI. 

Taking  the  logarithm  to  the  base  10,  we  have 

log^i  +  w(log  FI)  =  Iogp2  +  w(log  F2). 

_  log  pi  —  logj?2  _  log  2.5  _  0.3979  _  i  30 
~  log  F2  -  log  FI  ~  log  2    ~  0.3010  ~ 

6.  Entropy  changes.     Given  2  pounds  of  air  at  14.7  pounds  abso- 
lute pressure  and  32°  Fahrenheit,  find  and  plot  the  following  entropy 
changes. 

(a)  constant   pressure,  temperature   increases   to    736°  absolute, 
Fahrenheit. 
*  (5)  constant  volume,  temperature  increases  to  736°  absolute. 

(c)  constant  temperature,  pressure  decreases  to  (J)  (14.7)  pounds. 

(d)  what  is  the  entropy  change  in  Problem  5  ? 

Solution.    By  eq.  (38), 

(a)  02  -  0i  =  cp  loge  ^  =  (0.237)  ("2.303  log  ^1  =  0.0961. 

Since  this  change  in  entropy  is  per  pound,  the  total  change  is  0.192. 

.(6)  By  eq.  (37),        02-0t  =  cv  loge^  =  (0.169)  [2. 303  log  ^1  =0.0684. 

6\  L  49U 

The  total  change  in  two  pounds  of  air  is  then  0.137. 

1  For  all  of  the  problems  of  this  book  slide  rule  results  are  sufficiently  accurate.    The 
slide  rule  has  been  used  in  all  numerical  work  connected  with  these  problems. 


48 


THEftMOD  YN  AMIC  S 


(c)  By  eq.  (38),     02-0i  =  -(Cp-c,,)  log, -22=  (0.237 -0.169)  [2.30  log  1.5]  =0.0276. 

Pi 
Or  the  change  for  2  pounds  is  0.0552. 

These  changes  are  referred  to  in  Figure  14  by  the  marginal  letters  used  above. 

(<2)  By  eq.  (36),  02  -  0i  =cp  loge^  +  cv  log ^  =  2.30  [0.237  log  2  +  0.169  log  0.4] 

Vi  pi 

or  0.0090,  and  hence  for  2  pounds  0.018. 


900 
8OO 
700 
6OO 
500 
400 
300 
2OO 
100 
0 

e 

^ 

a 

^^ 

x"*^     ^ 

*^*^ 

^ 

c 

0 

O.O5       O.1O       O.15       O.2 

FIG.  14 

7.  Isothermal  expansion.  Three  cubic  feet  of  air  at  a  pressure 
of  500  pounds  absolute  and  a  temperature  of  450°  Fahrenheit  expand 
isothermally  to  a  volume  of  18  cubic  feet.  Find  the  heat  received. 
Find  the  work  done  per  pound  of  air. 

Solution.    pl  =  (500)  (144),         vi  =  3,         Oj.=  459.5  -f  450  =  910,        v2  =  18. 


By  eq.  (5), 
By  eq.  (18), 

By  eq.  (13), 


=  ^1^144=  (83.3)  (144).i 
(18) 


w=  (500)  (144)  (3)  [2-30  log?l  =  387,000  foot  pounds. 


&  =  AW=  887^000  =  498  B.  t.  u. 

778 


=  4.46  pounds. 


(53. 35)  (910) 
Therefore  work  per  pound  is  87,000  foot  pounds. 

8.    Adiabatic  expansion.       If  the  air  in  Problem  7  after  reaching 
the  volume  of  18  cubic  feet  is  allowed  to  expand  adiabatically  until 


1  It  is  convenient  to  solve  here  for  p2,  which  will  be  used  in  Problem  9. 


GASES 


the  temperature  is  200°  Fahrenheit,  what  is  the  amount  of  work 
done  per  pound  of  air? 

Solution.    Designate  the  state  after  adiabatic  expansion  by  the  subscript  3. 

p.2  =(83.3)  (144),         F2  =  18,         02  =  910,         03  =  459.5  +  200  =  660. 


By  eq.  (32), 
By  eq.  (19), 
Therefore 


Z2 
F2 


-  1 


(88.8)  044) 


K-l 


W  =  1S 


_e«r|  =  148,500  foot  pounds. 


The  work  per  pound  is      148*500  =  33,300  foot  pounds. 
4.46 

9.  Simultaneous  equations  in  p  and  v.  If  the  expansions  of  Prob- 
lems 7  and  8  are  two  transformations  in  a  Carnot  cycle,  find  the  p 
and  v  coordinates  of  the  four  intersections  of  isothermals  and  adia- 
batics.  What  is  the  efficiency  ?  Plot  the  cycle. 


500' 


400 


300 


200 


100 


10 


ZO  30 

FIG.  15 


Solution.     From  Problems  7  and  8, 

P!  =  500  Fi  =  3 

p2  =  83.3  Fo  =  18 

F3  =  40.2 

Find  p3  by  eq.  (22). 


40  cu.ft. 


01  =  910 

02  =  01 

03  =  660 


=  %£pa  =  f-^-V'4°(83.3)  =  27.1  pounds  per  square  inch. 
F3~  V40.27 


50 


THERMODYNAMICS 


Find  F4  by  solving  simultaneously  the  equations  for  the  adiabatic  through  state 
point  1  and  the  isothermal  through  state  point  3,  thus 


and    p4  =  Ms  =  (27.1)  (144)  (40  .2)  =  ^^  ^^  ^  Qr  16g  poundg 

The  values  of  j>,  F,  and  0  not  given  in  Problems  7  and  8  are  then 
pa  =  27.8 
^4=163  F4  =  6.7  04  =  03 

The  efficiency,  by  eq.  (25),  is  77  =  91°  ~  66°  =  27.5  percent. 

y  10 

The  plot  of  this  cycle  is  seen  in  Figure  15. 

10.  Pressure-volume  plot.  A  quantity  of  air  occupies  8  cubic  feet 
at  70°  Fahrenheit  and  under  an  atmospheric  pressure  of  15  pounds. 
It  is  compressed  adiabatically  to  a  volume  of  2  cubic  feet,  cooled  at 
constant  volume  to  its  original  temperature,  and  allowed  to  expand 
adiabatically  to  atmospheric  pressure.  It  then  heats  at  constant 
pressure  to  its  initial  temperature.  Draw  the  pressure  -volume  plot 
of  this  cycle. 


100 
90 
80 
7O 
60 
50 
40 
30 
20 
10 

P 

2 
\ 

\ 

\ 

v 

\ 

3 

\ 

\ 

\ 

\ 

\ 

V 

\ 

\ 

\ 

\ 

\ 

\ 

\ 

f\4 

\ 

• 

V 

234567  89 

FIG.  16 


Solution.    The  p-v  plot  is  shown  in  Figure  16,  which  has  been  drawn  by  the  use  of 
the  values  given  below. 


GASES 


51 


Given  px  VIK  =  p«VS,  where  p±  =  (15) (144),   Vl  =  8,  0X  =  460  +  70  =  530,  F2  =  2, 
=  1.40  ;  then 


—  104.6  pounds  per  square  inch. 


By  eq.  (31), 


Since  03  -530, 


=  922. 


=  r««2  =S^  104.6  -60. 
62          922 


During  adiabatic  expansion  to  atmospheric  pressure 

p  VK  =  ps  F3K  =  p4  Vf  and  p4  =  15. 


Hence 


and 


'4 


04  =  I  -A_V"(530)  =  358°  absolute,  or  -  102°  F. 
\5.35/ 


11.  Regenerative  cycle,  Sterling  engine.  The  Sterling  hot-air 
engine  works  upon  the  following  cycle  (as  shown  in  the  6-<f>  plot  of 
Fig.  17).  From  a  to  b  the  A  (j^H/. 

air  in  the  engine  expands 
isothermally  at  a  tempera-  1860° 
ture  0j,  from  b  to  c  this  air 
is  cooled  at  essentially  con- 
stant volume  to  the  tem- 
perature #3  and  pressure  jt?3, 
from  c  to  d  it  is  then  com- 
pressed practically  iso- 
thermally to  the  volume  v1 

and  a  pressure  p^,  from  d  \  (   'r^^ 

to  a  the  air  is  heated  at 
constant  volume  to  its 
initial  condition. 

This  is  accomplished  by 
the  use  of  a  displacement 
piston  D  in  conjunction 
with  a  working  piston  TPJ  549° 
as  is  shown  in  the  diagram 
of  Figure  18.  The'  dis- 
placement piston,  actuated 
by  a  rocker  arm,  moves  up 
its  full  stroke  during  da, 

while  the   working    piston  

is  essentially  at  rest.      It  FIG.  17 


52 


THERMODYNAMICS 


thus  forces  all  the  cool  air  at  temperature  03  through  the  regenerator 
R  into  the  lower  portion  beneath  which  there  is  a  fire  F.  The  re- 
generator consists 
merely  of  thin  metal 
plates  or  wires  which 
will  rapidly  absorb  or 
reject  heat.  The  air  in 
contact  now  with  the 
heating  surface  expands 
from  a  to  b  and  causes 
a  working  stroke  of  the 
piston  W-,  while  the  pis- 
ton D  remains  at  the 
top  of  its  stroke.  When 
W  has  reached  the  end 
of  its  stroke,  the  piston 
D  moves  quickly  down, 
thus  causing  the  air 
below  it  to  pass  through 
the  regenerator.  The 
air  passing  through  the 
regenerator  gives  up 
heat  and  then  passes  into  the  refrigerator  (7.  The  pressure  is 
thereby  reduced  from  b  to  c.  The  refrigerator  consists  of  a  set  of 
pipes  through  which  flows  cooling  water.  The  working  piston  W 
now  returns,  compressing  essentially  isothermally  at  the  temperature 
of  the  refrigerator  all  the  air  originally  contained  by  both  cylinders. 
This  last  transformation  is  represented  by  cd. 

Neglecting  clearance,  find  the  efficiency  for  such  an  engine  work- 
ing between  1400°  and  80°  Fahrenheit.  Assume  the  volume  of  the 
working  cylinder  to  be  0.9  cubic  foot  and  the  entire  volume,  exclu- 
sive of  the  working  cyclinder,  to  be  3  cubic  feet.  Compare  with  the 
Carnot  engine  in  efficiency.  Sketch  the  temperature-entropy  plot 
and  compare  with  the  plot  for  the  Carnot  cycle.  The  lowest  pres- 
sure is  16  pounds  per  square  inch. 

Solution.  Denote  by  the  subscripts  1,  2,  3,  and  4  the  state  points  a,  6,  c,  and  c?, 
respectively. 

81  =  62  =  460  +  1400  =  1860  03  =  64  =  460  +  80  =  540 

F!  =  F4  =  3  F2  =  F3  =  3  +  0.9  =  3.9 


FIG.  18 


GASES 


53 


Then 


e±      ^1860 

04 P4  ~  540 

02  „       I860 


By  eq.  (18), 


30  log          =  10.42  B.,, 


G-2-3  =  Oi-i,  since  they  are  constant  volume  changes  between  the  same  limiting 
temperatures. 


The  efficiency  is  then 
The  Carnot  efficiency  is 


,  = 


10.42 
1860  -  540 


=70.8  per  cent. 


1860 


=  70.8  per  cent. 


Discussion.  This  is  an  illustration  of  a  regenerative  cycle  where, 
although  the  changes  are  at  constant  volume  instead  of  isentropic,  the 
efficiency  is  the  same. 
There  are  other  cycles 
possible  which  have  the 

same  efficiency  as  the  c\;  ;  ;  ;  ;  ;  ;\b 
Carnot  for  the  same 
range  of  temperatures, 
but  none  has  a  higher 
efficiency,  as  is  explained 
in  Section  34. 

12.  Air  compressor 
with  and  without  clear- 
ance. An  air  compres- 
sor with  perfect  valves 
receives  air  at  the  at- 
mospheric pressure  of 
15  pounds  per  square 
inch  and  delivers  it  at  a 
constant  pressure  of  90 
pounds  per  square  inch. 
The  compression  follows 
the  equation  p  Vn  equal  to  a  constant  where  n  is  1.3  due  to  partial 
cooling.  Draw  the  p-v  plot  and  find  how  many  foot  pounds  of 


FIG.  19 


THEEMODYNAMICS 


work  are  required  for  each  cubic  foot  of  air  at  the  atmospheric  con- 
ditions under  which  the  condenser  is  working.     (#)  Neglect  clear- 
ance.   (6)  Allow  for  a 

P  ;  clearance  of   11.1   per 

cent  of  the  piston  dis- 
placement. 

Solution.  Representing 
the  total  piston  displacement 
and  clearance  by  FI,  the  vol- 
ume after  compression  by 
Fa,  the  clearance  volume  by 
F4,  and  the  volume  occu- 
pied by  the  clearance  air  at 
atmospheric  pressure  by  F3, 
Figures  19  and  20  represent 
respectively  the  cycle  and 
the  work  done  upon  the  air 
for  conditions  (a)  and  (6) 
above.  The  calculations  are 
tabulated  below  for  these 
two  cases.  It  is  important 
to  notice  that  the  net  work 
— -  done  on  the  air  is  only  that 
done  in  compressing  it  above 
the  15  pounds  at  which  it  is 
-  received. 


FIG.  20 


(a)  Without  clearance. 

Work  of  compression.    By  eq.  (19),      area  abhk  =  3650  V\ 

j_ 

Volume  expelled  at  90  Ib.  pressure,  F2  =  [ — \  ^  FI  =  0.251  FI 


(6)  With  clearance. 
area  abhk  =  3650  FI 


=  (0.261-  0.100) 
=  0.151  Fi 

area  besh  =  1960  PI 
5610  Fi 
area  afmk  =  1300  FI 

area  efms  =  1460  FI 
2850  Fi 
0.603  Fi 
4740 


Work  of  expulsion  (90)  (144)  (  F2  -  F4),  area  bcgh  =  3250  FI  • 
Total  work,  6900  Vl 

Work  done  by  atmosphere,  area  adgk  =  2160  FI 

Work  done  by  clearance  air.     (  F3  =0.397  Fi),     By  eq.  (15). 
Therefore  net  work  of  compressor,  4740  FI 

Cubic  feet  of  air  (at  atmospheric  pressure)  delivered,        V\ 
Foot  pounds  of  work  per  cubic  foot  delivered,  4740 

Discussion.  The  effect  of  the  clearance  in  a  frictionless  compressor 
is  not  to  increase  the  work  per  cubic  foot  of  air  delivered,  but  to  re- 
quire a  larger  piston  displacement  to  deliver  the  same  amount  of  air 
per  stroke. 

13.  Otto  gas  engine  cycle.  The  Otto,  or  de  Rochas,  cycle  for  a 
gas  engine  consists  thermodynamically,  in  the  ideal  case,  of  the  fol- 
lowing processes  :  (1)  adiabatic  compression  of  an  explosive  mixture 


GASES 


55 


of  air  and  gas;  (2)  heating  at  constant  volume  by  combustion; 
(3)  adiabatic  expansion  of  the  products  of  combustion  to  the  origi- 
nal volume  occupied  by  its  constituents;  (4)  cooling  at  constant 
volume. 

This  cycle  will  be  discussed  upon  the  following  assumptions : 
(1)  the  specific  volumes  of  the  constituents  will  be  considered  equal 
to  that  of  the  products  of  combustion  under  the  same  conditions  of 
temperature  and  pressure  ;  (2)  the  products  of  combustion  are  con- 
sidered to  follow  the  laws  of  perfect  gases  and  to  have  specific  heats 
in  the  same  ratio ;  (3)  the  density  of  the  gases  before  and  after  com- 
bustion are  considered  equal  to  that  of  dry  air  under  the  same  con- 
ditions of  pressure  and  temperature. 

The  p-v  plot  for  such  a  cycle  is  shown  in  Figure  21.  The  piston 
makes  four  strokes  for  each  working  stroke  or  explosion  During 
the  forward  stroke  ab  the  piston 
draws  into  the  cylinder  a  charge 
of  gas  and  air.  From  b  to  c  it 
compresses.  From  c  to  d  the  heat- 
ing takes  place  instantaneously, 
due  to  the  explosion  of  the  mix- 
ture. During  the  forward  stroke 
de  external  work  is  done.  The 
valve  then  Opens  at  e  and  the  gases 
exhaust,  the  pressure  and  tempera- 
ture dropping  to  that  of  the  at- 
mosphere or  the  "muffler."  In 
effect  it  is  as  if  the  valves  had  not 
opened  but  the  gas  had  been  cooled 
at  constant  volume  as  represented 
by  the  line  eb.  The  fourth  stroke 
now  expels  this  gas  against  atmos- 
pheric pressure  and  is  represented 
by  la. 

(a)  If  the  clearance  is  25  per  cent  of  the  piston  displacement,  find 
the  theoretical  thermal  efficiency. 

(5)  If  a  mixture  of  gas  and  air  which  will  develop  85  B.  t.  u.  by 
combustion  is  taken  into  the  cylinder  at  a  pressure  of  14  pounds  and 
a  temperature  of  60°  Fahrenheit,  find  the  temperatures  at  c  and  d 
and  the  pressures  at  e,  d,  and  e. 

(e)  Compare  the  efficiency  with  the  Carnot  efficiency  for  the  range 
of  temperatures  found  from  (6). 


v 


FIG.  21 


56  THERMODYNAMICS 

Solution,  (a)  All  the  heat  taken  in  is  along  cd  in  Figure  21.  Similarly,  the  heat 
rejected  is  along  eb.  The  calculation  will  be  made  for  one  pound  of  the  explosive  mix- 
ture. 


The  efficiency  is 

qi 

where  ^  =  cv(dd—  0C)  and  q2=  cv(0e  —  06). 

Hence  c/fr-g.)  -c,(ft- ft)  =         j. 


Since  de  and  cb  are  adiabatics,  they  must  differ  by  a  constant  quantity  of  entropy. 
Equating  then  the  difference  in  entropy  found  by  eq.  (37)  between  c  and  d  to  that 
found  between  e  and  &,  gives 

cv  log  —  =  cv  log  — ,  hence  —  =  —  • 
Therefore  6e  ~  0J>  =  -  • 

0d  —  0c        0c 

But  since  b  and  c  lie  on  the  same  adiabatic,  eq.  (32)  gives  2  =  (  SsV     .    Now  vc  is 

0C      \Vb/ 
clearance  and  vj>  is  piston  displacement  plus  clearance.     Hence 


_  1  _  0&  _  1  _  f  _  clearance  _  I*"1. 
ec  L 


clearance  +  piston  displacemen 
Substituting  for  the  clearance  25  per  cent  of  the  piston  displacement  gives 

r,  =  i  _  r   0.25^  -ii.*-i  =  l_  0  525  _  47 

L!  +  0.25J 

(6)  pa  zz  p5  =  14.          06  =  60  +  460  =  520.          ^  =  85  B.  t.  u. 


qi  =  c^^d  -  0C)  =  0.169(^  -  990)  =  85  B.  t.  u. 
Therefore    4  0d  -  1490. 

By  eq.  (32),  pc=/^Ypb  =(9.62)  (14)=  133. 

\vc/ 

Then  ^  =  ^  ;  therefore  jOd=  —  133  =  201. 

C      ec  990 


Byeq.(32),  ,. 

(c)  The  Carnot  efficiency  is 

gg^jg,  =  1490-  520  ^65percent 
0d  1490 

14.  Diesel  internal  combustion  cycle.  Gas  engines  operating  upon 
the  Diesel  cycle  compress  air  to  a  high  temperature,  as  shown  by 
bo  in  Figure  22.  The  temperature  is  sufficient  to  ignite  the  fuel, 
which  is  injected  from  c  to  d  at  a  rate  such  that  cd  is  essentially  an 
isothermal.  Considering  the  initial  compression  to  be  adiabatic  and 
from  15  pounds  atmospheric  pressure  and  60°  Fahrenheit  to  500 


GASES 


57 


pounds  gauge,  the  isothermal  to  be  for  15  per  cent  of  the  working 
stroke,  and  the  remainder  of  the  expansion  to  be  adiabatic,  solve  for 
the  following  :  (1)  the  temperature 
of  the  isothermal,  (2)  the  clearance 
in  per  cent  of  the  piston  displace- 
ment, (3)  the  B.t.  u.  supplied  by 
the  combustion  per  pound  of  work- 
ing substance,  (4)  the  thermal 
efficiency,  and  (5)  the  Carnot  effi- 
ciency for  the  same  range  of  tem- 
peratures. 


Solution. 
(1)      0C  = 


040 


=  1430°  abs.  or  970°  F. 
i 


(2)      v-c  = 


Pc 


;   therefore  vc=0.08*v 


or  vc  =  0.087  (vb  -vc). 

That  is,  the  clearance  is  8.7  per  cent  of  the  piston  displacement. 

(3)  Since  vd  =  0.15(v6  —  #c)  +  vc  =  2.73  vc, 

the  heat  supplied  along  the  isothermal  is  per  pound  of  mixture 


FIG.  22 


778 


98  B  t  u 


778     "  vc      778     °  vc 
Since  6e  =  (v^Y-led  =  p.73)(0.08^)lo.4Q1430  =  ^ 

\Vel  L  Vi  J 

the  heat  rejected  at  constant  volume  is 

q2  =  0.169(780  -  520)  =  44  B.  t.  u. 

=  55  per  cent. 


(4)  Therefore  the  efficiency  98  —  44  . 


•n  = 


(5)  The  Carnot  efficiency  is 


98 
1430  -  520  . 


1430 


'--  64  per  cent. 


15.  Air  refrigeration.  The  fact  that  air  falls  in  temperature 
during  an  adiabatic  expansion  is  made  use  of  in  the  design  of  air 
refrigerating  machines.  The  essential  features  of  such  a  machine  are 
shown  in  Figure  23.  The  air  is  received  by  the  working  cylinder 
W  from  the  room  which  is  to  be  kept  cool.  Let  its.  temperature 
and  pressure  be  Ba  and  pa  respectively.  The  air  is  compressed  adia- 


58 


THERMODYNAMICS 


3 


*J 


ft 


batically  as  shown  in  the  p-v  plot  of  Figure  24.  It  is  then  cooled  at 
constant  pressure  by  passing  it  through  a  coil  placed  in  a  tank  sup- 
plied with  cooling  water. 
After  this  it  expands  adia- 
batically  along  cd  as  shown 
in  the  plot,  and  does  work  in 
the  expansion  cylinder  E, 
which  returns  it  to  the  prime 
mover,  which  is  driving  the 
compressor.  The  tempera- 

Iture  of   the  air  is  now  con- 

[siderably  below   that  of  the 
I  -          •=.  ^   room  from  which  it  was  re- 

W  E  ceived.       It  then  enters  the 

room  and  is  heated  at  con- 
stant pressure. 

If  pa  is  14.7  pounds  abso- 
lute, pb  is  14  times  pa,  6a  is 
to  be  zero  degrees  Fahren- 
heit, how  much  heat  is  ab- 
stracted from  each  pound  of 
air  while  it  is  in  the  cooling 
Sketch  the  p-v  plot 


e 


FIG.  23 


room 


for  each  cylinder  and  the  0-<f> 
plot  for  the  entire  cycle.  Neglect  the  clearance.  Assume  that  the 
cooling  water  reduces  the  tem- 
perature of  the  air  to  95°  Fah- 
renheit. 


Solution. 

=  14  pa  =  206  pounds  per  square  inch. 

Calling  va  equal  to  «,  we  find  vj,  and 
by  eq.  (32)  and  (19)  as  follows  : 


=    *,.=  -.  0.154,, 


These  values  are  plotted  in  Figure 
25,  which  shows  the  work  of  the  work- 
ing cylinder  W. 


FIG.  24 


GASES 

Similarly,  since  6C  =  95  +  460  =  555  and  pc  =pb  andpd  = 

p 


59 


=       =      (0.154,)=  0.086,, 


=  f^.Yve  =  (14)0-7i(0.086  ,)  =  0.562,, 


w 


FIG.  25 

These  values  are  plotted  in  Figure  26,  which  shows  the  work  done  in  the  expansion 
cylinder  E. 


The  net  input  to  the  compressor  is  the  dotted  area  abed  in  Figure  26.     This  input 
will  be  found  by  parts  as  below  : 

area  abkr  =P^-paVa  _  (144)  /u<7)  (14)  (0.1  54)  -1  v  =  6100  v  foot  pounds< 
/c—  1  0.40 

area  bcok  =  pb(vb  —  vc)  =  (206)  (144)  (0.068)  v  =  1990  v  foot  pounds. 


area  cdho  = 


=  (144)  (14>7) 


.086)  -  (0.562) 


/c—  1  0.40 

area  tter/i  =  j9a(va  —  «d)  =  (144)  (14.7)  (0.438)  v  =  930  v  foot  pounds. 


60 


THERMODYNAMICS 


The  net  input  is  then  3760  foot  pounds  per  cubic  foot  of  entering  air. 

One  pound  of  air  at  the  admission  temperature  and  pressure  has  a  specific  volume  of 

$*-  12.39  =  11.6  cubic  feet. 


FIG.  27 

The  work  of  the  compressor  per  pound  of  air  pumped  is  then 
(3760)  (11.  6)  =  43,600  foot  pounds. 
The  B.  t.  u.  equivalent  of  this  work  is 


The  heat  abstracted  per  pound  by  the  room  it  is  desired  to  refrigerate  is 
cp(ed-ea)  =  0.237(460  -  260)  =  47.5  B.  t.  u. 

The  cooling  water  must  then  remove  the  sum  of  these  two  amounts  of  heat  or 
103  B.t.  u.  As  a  check  calculate  the  heat  abstracted  by  the  cooling  water.  It  is 
evidently  Cp(06-0c)  =0.237  (990-655)  =103  B.t.  u. 

The  temperature-entropy  plot  is  shown  in  Figure  27. 


CHAPTER   III 

WATER  AND  ITS   SATURATED  VAPOR 

44.  Phenomena  of  Vaporization.  In  accordance  with  the  kinetic 
theory  the  molecules  of  a  liquid  are  considered  to  be  in  motion  in  a 
manner  similar  to  the  molecules  of  a  gas,  except  for  the  fact  that 
the  average  free  path  which  a  molecule  travels  between  impacts  is 
very  much  less  in  a  liquid,  and  except  for  the  fact  that  the  molecules 
are  now  within  the  sphere  of  each  other's  influence  and  exert  large 
mutual  attractions.  The  continual  evaporation  which  takes  place 
from  the  surface  of  an  exposed  liquid  demands,  however,  the  gen- 
erally accepted  assumption,  that  occasional  molecules  possessing 
individually  kinetic  energies  higher  than  the  average  for  the  entire 
body  convert  a  portion  of  their  kinetic  energy  into  potential  energy 
by  breaking  away  from  the  surface  attraction  of  the  neighboring 
molecules  and  passing  into  the  space  above  the  liquid.  That  this 
free  evaporation  is  facilitated  by  an  increase  of  temperature,  that  is, 
by  an  increase  of  molecular  kinetic  energy,  is  a  fact  of  common 
knowledge  which  supports  this  theory. 

If  the  space  above  the  liquid  is  enclosed,  some  of  the  molecules^ 
which  have  escaped,  and  in  that  space  are  moving  essentially  after 
the  manner  of  gaseous  molecules,  will  in  the  course  of  their  motion 
impinge  upon  the  surface  of  the  liquid  and  once  more  become  a  part 
of  it.  When,  however,  there  are  as  many  molecules  entering  the 
liquid  in  this  way  as  leave  it  in  the  same  time,  there  is  a  so-called 
"  state  of  active  equilibrium,"  and  the  vapor  constituted  by  the  free 
molecules  is  said  to  be  saturated.  If  now  the  volume  above  the 
liquid,  which  these  free  molecules  occupy,  is  reduced,  there  will  be 
momentarily  a  larger  density,  and  hence  a  larger  number  of  impacts 
of  the  free  molecules  with  the  liquid,  and  hence  a  condensation  of 
the  vapor.  The  ultimate  result  will  be  a  reduction  of  the  number 
of  free  molecules  until  the  density  is  reduced  to  its  former  value 
and  the  state  of  active  equilibrium  is  again  attained.  Other  things 
being  equal,  the  density  of  a  vapor  in  contact  with  its  liquid  is  inde- 
pendent of  the  volume  which  the  vapor  is  free  to  occupy. 

61 


62  THERMODYNAMICS 

On  the  other  hand,  if  the  temperature  of  the  liquid  is  increased, 
the  equilibrium  will  be  momentarily  disturbed  by  the  escape  from  the 
liquid  of  a  larger  number  of  molecules  than  return  to  it  in  the  same 
time.  This  momentary  disturbance  lasts  until  a  new  stable  density 
of  greater  value  is  reached.  The  density  of  a  saturated  vapor  in 
contact  with  its  liquid  is  then  some  function  of  its  temperature. 

The  pressure  exerted  by  the  vapor  is  also  a  function  of  the  tem- 
perature, since  by  the  kinetic  theory  the  pressure  depends  upon  the 
energy  of  the  molecules  and  their  number  per  unit  volume.  The 
number  per  unit  volume  is  directly  proportional  to  the  density,  and 
hence  the  pressure  increases  with  a  twofold  rapidity  as  the  tempera- 
ture rises.  The  temperature  at  which  the  pressure  of  the  saturated 
vapor  would  be  equal  to  the  external  pressure  upon  the  surface  of  a 
given  body  of  liquid  is  called  the  boiling  temperature  of  the  liquid 
for  that  condition  of  pressure.  At  this  temperature  bubbles  of  the 
vapor  start  to  form  beneath  the  surface  of  the  liquid,  rise  to  the  top, 
and  burst  into  vapor.  After  boiling  commences  no  further  increase 
in  the  temperature  (for  constant  pressure)  is  possible  until  all  the 
liquid  has  been  converted  into  vapor. 

Water  vapor,  or  steam,  as  it  is  usually  called,  is  colorless  and  trans- 
parent. The  cloudy  white  form  it  frequently  takes  is  due  to  the 
condensation  of  part  or  all  of  the  original  steam  into  minute  drops 
of  water.  In  this  case  the  steam  is  called  "wet"  because  of  the 
mixture  with  it  of  water  particles.  Steam  free  from  moisture, 
whether  due  to  condensation,  entrained  water,  or  other  causes,  is 
called  "dry." 

The  temperature  and  pressure  of  water  vapor  are  always  the  same 
as  the  temperature  and  pressure  of  the  water  with  which  it  is  in 
contact.1  If,  however,  the  saturated  vapor  is  removed  from  contact 
with  its  liquid,  its  temperature  may  be  raised  above  the  original  boil- 
ing point.  The  steam  is  then  said  to  be  superheated.  If  the  pres- 
sure of  the  superheated  steam  is  kept  constant,  its  volume  increases 
with  the  temperature ;  and  also,  if  its  volume  is  kept  constant,  the 
pressure  increases  with  increased  temperature.  In  steam  engineer- 
ing practice  the  steam  is  superheated  in  its  passage  from  the  boiler 
to  the  engine  as  the  result  of  passing  through  coils  of  pipe  in  the  flue 
of  the  boiler  or  as  the  result  of  some  similar  arrangement.  Its  vol- 
ume per  pound,  or  specific  volume,  then  increases,  and  this  increase 
is  cared  for  at  once  by  the  motion  of  the  piston  of  the  engine.  The 

1  Except  for  differences  beneath  the  surface  of  the  liquid  due  to  the  hydrostatic  pressure. 


WATER  VAPOR 


63 


effect  is  the  same  as  if  the  steam  were  entirely  separated  from  the 
water  from  which  it  is  produced. 

45.  Specific  Heat  of  Water.  In  order  to  convert  a  pound  of  water 
into  steam,  it  is  necessary,  tirst,  to  raise  the  temperature  of  the  water 
to  that  corresponding  to  the  boiling  point  for  the  given  pressure, 
and,  second,  to  vaporize  this  water.  In  each  case  the  necessary  heat 
may  be  expressed  by  an  equation  of  the  form  dQ  =  AdiS-\-  Adl+  AdW, 
as  in  Section  8.  In  raising  the  temperature  of  the  water  there  is 
produced  a  small  increase  in  its  volume.  For  ordinary  engineering 
purposes  this  is  so  small  that  the  terms  Adi  and  AdW  may  be  neg- 
lected. This  necessary  increase  in  the  heat  of  the  liquid  will  be 
represented  by  the  symbols  q%  —  q^  the  significance  of  which  will 
appear  more  fully  later  in  this  section. 

This  increase  in  the  heat  of  the  liquid  may  be  found  approxi- 
mately by  assuming  the  specific  heat  to  be  constant  and  of  average 
value  unity.  Then  the  heat,  ^""^i1  required  to  change  one  pound 
of  water  from  a  temperature  of  ^  to  the  boiling  temperature  £2  is 
expressed  as  q2  —  ql  =  t2  —  tr  It  is  in  general  desirable  to  know  this 
necessary  heat  more  accurately,  and  this  requires  a  knowledge  of 
the  specific  heat  of  water  at  various  temperatures.  The  curve  of 
Figure  28  shows  the  variations  of  the  value  of  the  specific  heat  of 


1.0400 


1.0300 


1.0100 


1.0000 


0.9900 


50 


150  200  250 

Degrees  Fahrenheit 

FIG.  28 


3OO 


35O 


4OO 


water,  taking  the  specific  heat  as  unity  for  the  change 
Fahrenheit.  From  such  a  curve,  accurately  constructed  on  a  large 
scale,  it  would  be  possible  to  obtain  the  value  of  the  specific  heat  for 
each  degree  and  then  by  a  summation  of  the  form  ^cdt  to  obtain 

fc  -  ?r 

It  is  more  convenient,  however,  to  use  some  set  of  tables,  in  which 

are  tabulated  the  results  of  such  a  summation  from  32°  Fahrenheit, 
as  a  lower  limit,  to  t°  as  an  upper  limit,  where  t  has  for  values  all 


64 


THERMODYNAMICS 


the  whole  degrees  within  the  ordinary  range  of  engineering  work. 
The  heat  which  is  required  to  raise  one  pound  of  water  from  32°  to 
t£  will  bQ  represented  by  qr  Similarly,  the  heat  in  B.  t.  u.  required 
to  raise  one  pound  of  water  from  32°  to  £2°  is  q2.  The  heat  required 
to  raise  one  pound  of  water  from  ^  to  £2  is  then  q%  —  q1  B.  t.  u.  Such 
a  table  of  values  for  q  is  illustrated  in  Table  3,  on  page  130,  for  a  few 
chosen  values.  Certain  other  constants  for  steam,  which  will  be 
discussed  later,  enter  into  this  table. 

The  plot  of  Figure  29  shows  the  general  nature  of  the  relation 
between  pressure  and  temperature  for  water  vapor.       It  gives,  in 


7400 


100 


180  160  140  120  100  SO 

Pounds  per  square  inch 

FIG.  29 


20 


other  words,  the  boiling  point  temperature  corresponding  to  a 
given  pressure.  There  is  to  be  noticed  at  the  higher  temperatures 
the  rapid  increase  of  pressure  mentioned  in  Section  44.  In  Table  3, 
the  value  of  p  corresponding  to  any  given  temperature  t  is  the  pres- 
sure in  pounds  per  square  inch  for  which  t  is  the  boiling  point. 

46.  Heat  of  Vaporization.  The  heat  required  to  vaporize  one 
pound  of  water  may  be  considered  in  three  parts.  The  first  of 
these,  namely,  q%  —  qv  which  is  required  to  raise  the  temperature  of 
the  water  to  the  boiling  point,  has  already  been  discussed.  The 
remaining  two  have  been  considered  partially  in  the  illustration 
given  on  page  7.  They  are  p,  the  internal  latent  heat  required  to 
separate  the  molecules  against  their  mutual  attractions,  and  A.pu, 
the  external  latent  heat  or  heat  equivalent  of  the  external  work 
which  must  be  done  against  the  pressure  p  to  cause  the  increase,  u, 
in  the  volume.  This  increase  u  is  s  —  <r,  where  s  represents  the  spe- 
cific volume  of  one  pound  of  dry  steam  at  the  pressure  p,  and  a- 
represents  the  volume  of  one  pound  of  water. 


WATER   VAPOE  ,  65 

The  total  heat  required  to  raise  one    pound  of  water   from    32° 
Fahrenheit  to  a  temperature  t  and  there  to  vaporize  all  of  it  is  then 

\  =  q  +  p  +  Apu,        (43) 

where  u  =  s  —  cr.         (44) 

The  letter  X  will  be  used  hereafter  as  a  general  symbol  for  the  total 
heat  contents  of  a  pound  of  water  (in  whatever  form  it  may  be)  as 
measured  by  the  heat  in  B.  t.  u.  required  to  change  the  water  from 
32°  Fahrenheit  to  the  condition  for  which  X  represents  the  heat  con- 
tents. Also  the  letter  r  will  be  used  to  denote  the  total  latent  heat 
of  vaporization.  Thus 

r  =  p  +  Apu .     .     (45) 

In  Figure  30  is  shown  the  general  relation  of  q,  p,  Apu,  X,  and  8 
for  steam  at  various  temperatures.  It  is  to  be  noticed  that  the  term 
Apu,  which  is  represented  in  the  figure  by  the  portion  of  the  ordi- 
nate  between  the  curve  of  p,  the  "  internal  latent  heat,"  and  r,  the 
total  "  latent  heat,"  is  almost  constant.  Since  the  specific  volume  of 
a  pound  of  water  is  always  approximately  0.016  cubic  foot,  it  may 
for  many  purposes  be  entirely  neglected  as  compared  with  the  spe- 
cific volume  s  of  dry  steam.  Hence  the  product  Apu  is  practically 
equal  to  Aps.  This  relation  between  p  and  s  has  been  expressed  in 
empirical  formulas  for  calculating  the  specific  volume  of  dry  steam 
for  any  pressure.  Thus  by  Zeuner  as 

pslM*  =  479,     ,.-.     ; (46) 

and  by  Rankine  as 

JEW**  =  482,      .     .     .     .     .     .   ''A  (47) 

These  equations  obviously  make  allowance  for  the  fact  that  Apu  is 
not  exactly  constant  but  increases  somewhat  at  higher  pressures. 

The  values  for  p,  r,  p,  Apu,  s,  and  7,  the  density  or  mass  per  cubic 
foot,  for  selected  values  of  t  are  given  in  Table  3  on  page  130.  *  It 
is  to  be  remembered  that  these  values  of  r,  p,  Apu,  as  well  as  q,  are 
all  in  B.  t.  u.  per  pound.  The  specific  volume  s  is  in  cubic  feet  per 
pound.  The  pressure  p  is  in  pounds  weight  per  square  foot,  but 
since  pressures  are  usually  measured  by  gauges  graduated  in  pounds 

1  A  convenient  set  of  steam  tables  is  the  work  of  Professor  Peabody  of  the  Massachusetts 
Institute  of  Technology.  It  is  published  under  the  title  "  Steam  and  Entropy  Tables  "  by 
Wiley  and  Sons.  The  values  used  in  this  text  and  included  in  Table  3  are  taken  from  the 
8th  edition  by  the  kind  permission  of  the  author  and  the  publishers.  The  symbols  used  in 
Table  3  are  those  of  Professor  Peabody  except  in  the  case  of  entropy,  which  is  considered 
later  in  Sections  49  and  50.  The  tables  compiled  by  Professors  Marks  and  Davis  of 
Harvard  University  also  embody  in  a  most  convenient  form  the  results  of  the  recent  inves- 
tigations (Longmans,  1908). 


66 


THERMODYNAMICS 


per  square  inch,  the  values  of  the  table  in  the  column  headed  by  p 
are  so  expressed  and  must  be  multiplied  by  a  factor  of  144  before 
being  used  in  any  numerical  calculations. 

It  is  to  be  noticed  also  from  Figure  30  that  />,  the  internal  latent 
heat  of  vaporization,  decreases  with  the  volume.     This  is  in  accord- 


12OO 


1100 


1000 


900 


800 


700 


250 


300 
Degrees  Fahrenheit 

FIG.  30 


350 


4OO 


ance  with  the  kinetic  theory,  for  when  occupying  a  smaller  volume 
the  molecules  are  closer  together  and  less  work  has  been  done  in 
producing  vaporization  against  their  mutual  attractions. 

47.    Quality  of  Steam.     Consider  a  pound  of  water  at  32°  Fahren- 
heit, and  the  results  of  raising  its  temperature  to  t°  and  there  evap- 


WATER   VAPOE  67 

orating  against  the  corresponding  pressure  of  p  pounds,  not  the 
entire  pound  of  water  as  the  problem  has  previously  been  consid- 
ered, but  a  fraction  of  a  pound,  say  x.  Then  obviously  the  funda- 
mental equations  given  below  on  the  left  hand  must  be  restated  as 
shown  on  the  right  hand. 

\  =  q  +  r  =  q  +  p  +  Apu.  .  .  .  (43)  \  =  q  +  xr  =  q  +  xp  +  xApu.  .  (48) 
v  =  s  =  u  +  <r (44)  v  =  xs  +  (I  —  x~)o-  =  xu  +  o-.  .  .  (49) 

In  general,  if  from  any  cause  the  steam  considered  is  not  dry  but 
has  mixed  with  it  water,  and  the  mixture  is  x  parts  dry  steam  and 
(1  —  x)  parts  water,  then  x  is  defined  as  the  "  quality,"  and  the 
modified  equations  given  above  must  be  used.  The  quality  x  is 
usually  expressed  in  per  cent,  but  obviously  it  must  enter  into  these 
equations  as  a  decimal.  The  amount  of  water  present,  (1  —  x)  parts, 
is  expressed  as  a  per  cent  and  called  the  " priming." 

48.  Intrinsic  Energy  of  Steam.     In  the  formation  of  one  pound  of 
dry  steam  at  a  pressure  p,  it  is  to  be  noted  carefully  that  the  mole- 
cules of  the  forming  steam  act  as  direct  agents  by  which  a  certain 
amount  of  mechanical  work,  pu,  is  done  at  the  expense  of  Apu  heat 
units  supplied  by  the  fire.     At  no  time  do  these  pu  units  constitute 
part  of  the  intrinsic  energy  of  the  steam.     Therefore,  in  the  equa- 
tion \  =  q  +  p+Apu,  which  is  in  the  form  of  Q  =  A(dS  +  dI-\-  dW), 
the  term  q  -f-  p  represents  the  intrinsic  energy  of  the  steam  and  cor- 
responds to  AdS+  Adi.     Representing  by  e  the  intrinsic  energy  of  a 
pound  of  steam  and  water  mixture,  we  have,  if  x  is  unity, 

Ae  =  q  +  p (50) 

Or,  more  generally,  for  the  intrinsic  energy  of  one  pound  of  a  mix- 
ture of  quality  x,                       A  xr-ix 
J                           Ae  =  q  +  xp (51) 

49.  Entropy  of  the  Liquid.     By  definition,  the  entropy  of  a  body 
in  state  2  is  the  integral  J    —3,  taken  along  any  reversible  path  from 

state  point  1,  arbitrarily  chosen  as  a  zero  of  entropy.  By  conven- 
tion for  steam  the  point  1  is  taken  as  the  entropy  of  a  pound  of 
water  at  32°  Fahrenheit.  The  heating  of  water  is  a  reversible  pro- 
cess, and  the  integral  representing  the  entropy  of  one  pound,  at  any 
temperature  t°  Fahrenheit  or  6°  Fahrenheit  on  the  absolute  scale, 
may  then  be  written  as 


.....     (52) 

0=491.5 


68  THERMODYNAMICS 

As  an  illustration,  the  entropy  of  water  at  33°  Fahrenheit  will  be 
calculated.     Thus 

,       492.5 


By  a  proper  allowance  for  the  varying  value  of  the  specific  heat  c, 
the  entropy  of  water  may  be  calculated  in  this  way  for  various  tem- 
peratures. Values  so  calculated  are  given  in  Table  3  in  the  column 
headed  fa.  This  symbol  will  be  used  hereafter  to  represent  the 
entropy  of  one  pound  of  water. 

50.  Entropy  of  Vaporization.  In  a  somewhat  similar  manner, 
since  vaporization  is  a  reversible  process,  the  increase  of  entropy 
fa  due  to  vaporization  may  be  found.  Thus 


(53) 


where  the  value  of  6  is  the  absolute  temperature  at  which  vaporiza- 
tion occurs,  and  r  is  the  heat,    I  d§.     As  an  illustration,  the  entropy 

of  vaporization  for  212°  Fahrenheit  will  be  calculated.     Thus,  sub- 
stituting for  r  from  Table  3, 

.  969.7 

*'  =  459.5  +  212  = 

The  symbol  fa  will  be  used  to  represent  the  increase  in  entropy  per 
pound  of  steam  due  to  vaporization. 

The  total  entropy  of  a  pound  of  dry  steam  is  then 


(54) 


If  the  quality  is  not  unity,  but  is  #,  then  the  total  entropy  of  a 

pound  of  the  mixture  is  /^Jt 

<t>  =  <l>w  +  xfa  ......     .>-•.     (55) 

51.  Temperature-Entropy  Chart  for  Steam.  If  the  temperature- 
entropy  diagram  for  the  change  in  entropy  of  one  pound  of  water 
from  32°  Fahrenheit  to  some  temperature  ^°  is  plotted,  it  will  appear 
as  shown  in  Figure  31  by  the  line  Fab.  If  the  diagram  is  constructed 
for  several  higher  temperatures,  as  £2,  £3,  and  so  on,  it  will  be  found 
that  for  these  temperatures  the  changes  in  entropy  from  32°  are 
represented  by  lines  Fed,  Fef,  and  so  on.  Such  a  temperature-entropy 
diagram  for  one  pound  of  water  when  constructed  for  a  sufficient 
number  of  temperatures  is  usually  known  as  a  "  temperature-entropy 
chart  for  steam."  It  is  of  considerable  value  in  the  solution  of 
problems  in  steam  turbine  study  and  of  interest,  although  probably 


WATER  VAPOR 


69 


32°  F 


FIG.  31 


70  THERMODYNAMICS 

of  lesser  value,  in  the  study  of  the  steam  engine.  Such  a  tempera- 
ture-entropy chart  is  given  in  part  on  page  86. 

The  curve  'Face  of  Figure  31  is  called  the  "  water  line,"  and  the 
locus  of  the  points  like  5,  d,  and/,  is  the  "dry  steam  line"  or  "line 
of  constant  steam  weight." 

Since  the  entropy  of  a  mixture  is  <f>w  +  x'<j>v  and  the  length  of  the 
lines  ab,  cd,  and  ef  represents  values  of  <f>v,  the  entropy  of  any  mix- 
ture may  be  easily  determined  from  this  chart.  Thus  if  the  entropy 
of  a  mixture  of  quality  x  at  a  temperature  t%  is  desired,  we  proceed  as 
follows.  Divide  the  line  cd  into  two  parts  by  a  point  g  so  located 
that  eg jcd  shall  equal  x.  Then  obviously  the  length  of  the  line  eg  is 
x<f>v  and  the  total  entropy  of  the  mixture  is  represented  by  the  line 
Off.  For  convenience  in  making  such  interpolations  on  the  chart 
a  series  of  lines  like  hgj  may  be  drawn,  indicating  all  the  points 
of  the  same  quality  for  values  of  x  differing  by  10  per  cent,  as  is 
done  in  Figure  32.  (In  the  chart  on  page  86  such  lines  have  been 
drawn  for  differences  of  one  per  cent.)  These  lines  of  constant 
quality  are  usually  called  "a?  lines." 

Since  the  temperature  is  constant  during  vaporization,  if  the 
pressure  is  constant,  the  lines  ab,  cd,  and  so  on  represent  constant 
pressure  lines,  the  values  of  which  correspond  to  the  temperatures 
tv  £2,  and  so  on. 

Lines  of  constant  volume  showing  the  entropy  of  all  states  of  a 
mixture  of  steam  and  water  which  occupy  per  pound  the  same 
volume  may  be  drawn  as  follows.  Continue  the  axis  of  tempera- 
tures below  the  entropy  axis  and  let  it  represent  to  a  convenient 
scale  an  axis  of  volumes.  Draw  a  line  parallel  to  the  entropy  axis 
and  below  it  by  an  amount  equal  to  cr,  the  specific  volume  of  water. 
In  Figure  31  the  distance  cr  is  made  disproportionately  large  for 
clearness.  The  letters  af,  c',  and  e'  refer  to  the  intersections  of  the 
dotted  vertical  projection  lines  with  this  constant  volume  line  through 
cr.  Then  for  any  pressure  and  temperature  as  ^  project  normal  to 
the  entropy  axis  the  line  aaf  to  intersect  the  constant  volume  line  cr 
in  a',  and  the  line  W  to  intersect  the -line  s-fi' ,  which  is  drawn 
through  «j,  the  point  on  the  axis  of  volumes  which  corresponds  to 
the  specific  volume  of  dry  steam  at  the  temperature  tr  The  line  a'b1 
shows  then  the  increase  of  entropy  with  increasing  volume  due  to 
vaporization  at  this  temperature.  In  a  similar  manner,  other  lines, 
showing  the  relations  of  volume  and  entropy  for  the  temperatures  t2, 
ty  and  so  on,  may  be  drawn.  These  are  represented  by  cfd'  and  e'f. 
For  any  volume  as  k  the  line  of  constant  volume  kkfff  intersects  the 


WATER   VAPOR 


71 


lines  a'bf,  c'd',  and  e'f  in  points  &',  k",  and  k"f,  the  abscissas  of  which 
represent  the  entropy  of  mixtures  of  constant  volume  k  but  of  differ- 
ent temperatures.  The  corresponding  temperatures  may  be  found  by 
projecting  the  points  &',  k",  and  k1"  back  onto  the  lines  a£>,  cd,  and  ef, 
respectively.  The  ordinates  of  these  projections,  namely  k^  &2,  and 
&3,  represent  the  temperatures  for  mixtures  of  volume  k.  The  line 
k^B  is  then  on  the  temperature-entropy  plot  a  line  of  constant 
volume,  and  similarly,  the  lines  a'b',  c'df,  and  e'f  on  the  volume- 
entropy  plot  are  lines  of  constant  pressure  and  temperature.  On 
the  chart  of  page  86  several  constant  volume  lines  are  represented. 

A  family  of  curves  for  constant  total  heat  contents  might  also  be 
constructed  through  all  the  points  for  which  X  =  q  -f-  xr  —  constant, 
for  different  values  of 
the  parameter  X.  These 
lines  of  constant  heat 
contents  are  of  especial 
value  in  the  solution  by 
the  use  of  a  &-<]>  chart 
of  problems  on  the  flow 
of  steam  through  noz- 
zles. There  is  unfortu- 
nately no  graphical 
method  of  construct- 
ing constant  heat  con- 
tent curves,  and  the 
calculation  and  plotting 
of  separate  points  is 
necessarily  tedious. 
Constant  heat  curves 
are  shown  in  the  chart 
on  page  86.  Similarly, 
curves  for  constant  in- 
trinsic energy  might 
be  plotted  in  accordance  with  the  relation  e  =  q  +  xp  =  constant.  The 
general  appearance  of  these  curves  is  similar  to  those  for  constant 
heat  contents,  but  slightly  different  in  slope.1 

52.  Isentropic  Transformation  of  a  Mixture.  From  a  study  of  the 
temperature-entropy  diagram  of  Figure  32  it  is  evident  that  if  an 

1  For  a  method  of  determining  graphically  the  intrinsic  energy  corresponding  to  a  desired 
point  on  the  0-0  chart,  the  reader  is  referred  to  Berry,  "  Temperature  Entropy  Diagram,'* 
1st  edition,  pp.  28-29.  For  the  converse  problem  there  is  no  graphical  solution. 


FIG.  32 


72  THERMODYNAMICS 

isentropic  compression  abc  be  made,  starting  with  a  mixture  the  state 
of  which  is  represented  by  #,  as  successive  lines  of  smaller  volume 
are  intersected  (volume  lines  are  not  shown  in  the  figure),  so  also 
are  met  lines  of  greater  quality  Or  higher  values  of  x.  That  is,  an 
adiabatic  compression  of  steam  near  its  saturation  point  causes 
reevaporation,  the  heat  for  which  is  obtained  at  the  expense  of  the 
external  work  done  during  compression.  When  the  point  c  is 
reached,  further  isentropic  compression  results  in  superheating  the 
steam,  as  was  first  announced  by  Cazin.  Conversely,  starting  from 
6?,  an  isentropic  expansion  results  in  a  condensation  or  lowering  of 
the  quality  #,  as  was  first  stated  by  Him.  That  is,  the  external 
work  is  now  done  by  the  expanding  steam  at  the  expense  of  the 
heat  of  vaporization  of  a  part  of  it. 

From  the  same  figure  it  will  be  noticed  that  the  x  lines  for  those 
qualities  less  than  0.5  in  general  slope  to  the  left,  while  the  slope 
is  toward  the  right  for  those  qualities  greater  than  about  0.5.  Hence 
for  the  isentropic  line  efg,  when  the  mixture  is  near  its  liquid  condi- 
tion, it  is  evident  that  compression  results  in  condensation,  and,  vice 
versa,  expansion  results  in  reevaporation,  as  was  stated  in  Section  44. 

An  isentropic  line  as  mn  of  Figure  32  if  produced  will  in  general 
cut  an  x  line  in  two  points.  Hence  for  each  x  line  an  isentropic  may 
be  drawn  tangent  to  it  at  some  point,  as  o.  If  the  line  o'o"  be 
drawn  through  all  these  possible  points  of  tangency,  its  general  ap- 
pearance will  be  as  shown  in  the  figure.  Such  a  line  is  called  the 
"  zero  line  "  (because  it  represents  states  for  which  the  specific  heat 
may  be  shown  to  be  zero).1 

The  changes  in  the  state  of  a  mixture  undergoing  an  isentropic 
compression  may  then  be  expressed  in  a  more  rigorous  form  by  say- 
ing that  for  values  of  x  greater  than  the  zero  line  values  there  is  a 
reevaporation,  and  for  values  less  than  the  zero  line  values  there  is  a 
condensation.  Conversely,  for  isentropic  expansion  there  will  be 
condensation  (or  reevaporation)  for  values  of  x  greater  than  (or 
less  than)  the  zero  line  values. 

Since  for  an  isentropic  change  the  external  work  performed  by  the 
steam  is  at  the  expense  of  the  intrinsic  energy,  the  work  w  per  pound 
of  a  steam  mixture  undergoing  a  reversible  adiabatic  expansion 
from  state  1  to  state  2  is 

W  =  e1-e2  =  q1  +  x1p1-q2-x2p2.      ....      (56)' 
1  Berry,  "  Temperature  Entropy  Diagram,"  pp.  25-26. 


WATER  VAPOR  73 

53.  Specific  Heat  of  Steam.  The  specific  heat  of  dry  steam  may 
be  found  for  any  of  the  following  conditions :  (a)  superheating 
under  constant  pressure,  (£>)  superheating  at  a  constant  volume, 
(c)  for  changes  of  both  pressure  and  volume  so  related  that  the  whole 
mass  remains  dry  saturated  steam,  that  is,  for  changes  such  that  the 
temperature  and  entropy  are  related  as  in  the  dry  steam  line  fb  of  Fig- 
ure 31,  page  69.  For  constant  pressure  the  value  of  0.48  may  be  used 
for  temperatures  near  that  of  saturation  ;  for  higher  temperatures, 
values  are  given  in  the  Table  on  page  93.  For  constant  volume, 
Regnault  gave  0.346,  but  this  determination  is  not  entirely  reliable. 

For  condition  (<?)  the  specific  heat  is  negative  for  all  ordinary  tem- 
peratures and  pressures.  The  physical  significance  of  this  fact  may 
best  be  shown  by  an  example.  Thus,  suppose  it  is  desired  to  raise 
the  temperature  of  one  pound  of  steam  from  212°  to  213°  Fahrenheit 
in  accordance  with  condition  (<?)  above.  This  may  be  done  by  com- 
pressing the  steam  under  a  pressure  which  increases  according  to 
the  relation  plotted  in  Figure  29  of  page  64.  The  specific  volume 
decreases  with  this  increased  temperature  and  pressure.  The  exter- 
nal work  done  may  be  taken  as  approximately  equal  to  the  product 
of  the  average  pressure  and  the  decrease  in  volume.  This  is  work 
done  upon  the  steam  by  some  external  source.  The  change  in  total 
heat  contents  represents  the  amount  of  heat  which  must  be  added 
from  some  external  source.  The  values  of  jo,  q,  a,  and  r  for  212°  and 
for  213°  Fahrenheit  are  given  below. 

213° 

^  =  14.99 
*  =  26.29 


Hence,  0.4  B.  t.  u.  must  be  supplied  from  some  external  source  to 
change  from  212°  to  213°.     But  an  amount  of  external  work 

14.70+14.99  144(26.78  -  26.29)  =  1020  ft.  Ib.  =  1.35  B.  t.  u. 

2*      '*> 

has  been  done  upon  tlfe  steam.  .Now,  to  maintain  the  required  con- 
ditions, external  work  to  the  value  of  1.35  B.  t.  u.  has  been  done, 
but  only  0.4  B.  t.  u.  are  needed  to  change  the  steam  from  212°  to 
213°.  Hence,  unless  this  excess  of  (1.35  —  0.4)  obtained  from  the 
external  work  is  to  go  into  superheating,  it  must  be  subtracted. 


212° 

213° 

212° 

q=  180.3 

q=  181.3 

^  =  14.70 

r=  969.7 

r=  969.1 

*=26.78 

X  =  1150.0 

\=  1150.4 

74  THERMODYNAMICS 

This  amount  of  heat  per  pound  or  0.95  B.  t.  u.  must  therefore  be 
conducted  away,  and  hence  the  specific  heat  is  negative.1 

54.  Resume  of  Equations  for  a  Mixture  of  Steam  and  Water. 
The  following  equations  refer  to  a  pound  of  mixture,  and  are  num- 
bered as  in  the  preceding  sections. 

Total  Heat  Contents. 


Apu  =  Ae  +  Apu 
or  X  =  q  +  xr  =  q  +  xp  -f  xApu  =  Ae  +  xApu  .     .     (48) 


£;     or  $=£.  +  *$..   .     .     (55) 

U 


Total  Entropy. 

f«£*£ 

Total  Intrinsic  Energy. 

Ae=q  +  p,     oi'Ae  =  q  +  xp.        .     .     .     .      (51) 

Specific  Volume.     For  water  cr=  0.016.          For  steam  (dry)  s. 

v  =  xs  +  (1  —  X)G  =  xu  +  cr.        .      .     .     .     (49) 
Increase  of  Volume.         u  =  s  —  a-,     or  xu  .......      (44) 

Constant  Heat  Transformation. 

\  =  X2,     or  q1  -f  x^  =  q2  +  x2ry  .  .  .  •  -.     .     (48) 

Constant  Energy  Transformation. 
ej  =  e2,     or 

Isothermal  Transformation. 
n  n 

Constant  Pressure. 


} 

\  q,  6,  and  p  are  constant. 

}    J 


Isentropic  Transformation.  1 
Reversible  Adiabatic.  ) 

*i  =  ^2  =  ^1  +  a;  A,  =  *»1+a?2*v  •     '•     •     * 

w  =  ea-€1  =  -(92  -^1  +  ^2-^1)-    •     •     •     (56) 
J\. 

Irreversible  Adiabatic  Change. 

dq=0,  but  <ty=jfcO. 

1  This  method  is  of  course  unsatisfactory  for  the  purpose  of  determining  this  specific 
heat  because  of  the  large  errors  introduced  in  a  difference  between  two  quantities  where 
the  difference  is  of  a  magnitude  not  much  greater  than  the  probable  error  in  the  quantities 
subtracted.  For  the  method  used  by  Clausius  see  Preston,  "  Heat,"  pp.  655-658.  For  212° 
Fahrenheit  Clausius  gives  1.13.  The  method  used  above  is  selected  because  it  is  free  from 
differential  notation,  and  emphasizes  the  physical  relations  involved. 


WATER   VAPOR 


75 


PROBLEMS   AND   SOLUTIONS:    SATURATED   WATER 

VAPOR 

16.  Quality.  How  much  heat  has  been  required  to  change  one 
pound  of  water  at  60°  Fahrenheit  into  a  mixture  of  steam  and  water 
occupying  2  cubic  feet  at  a  pressure  of  130  pounds  per  square  inch 
absolute  ? 


Solution.     From  Table  3, 

For  130  pounds 
q2  =  318.6 

v2  =  872.1 
s2  =  3.451 
By  eq.  (49),     v2=2  =  X2s2  +  (1  - 


For  60°  F. 

i  =  28.1 


or  x2  = 


=  0.577. 


By  eq.  (48),  X2  -  \i  = 


-ql  =  318.6  +(0.577)  (872.1)-  28.1  =  793B.t.u. 


17.  Intrinsic   energy.     Find    the    intrinsic    energy,    volume,    and 
entropy  of  a  mixture  of  5  pounds  of  steam  and  water  which  is  80  per 
cent  steam,  if  the  pressure  is  120  pounds  absolute  per  square  inch. 
How  much  heat  must  be  added  to  make  the  quality  unity  at  this 
pressure  ? 

Solution.     From  Table  3,  for  120  pounds, 

q  =  312.3,     r  =  876.9,     p  =  794.2,     0W  =  0.4922,     fo  =  1.0951,     s  =  3.723. 

Given  x  =  0.80. 

By  eq.  (51),  energy  per  pound  =  778  [312  +  (0.80) (794. 2)]  =  736,000  foot  pounds, 

or  3,680,000  foot  pounds  for  5  pounds. 

By  eq.  (55),  entropy  per  pound  =  0.4922  +  (0.80)  (1.0951)  =  1.3682, 
or  6.841  for  5  pounds. 

By  eq.  (48),  total  heat  per  pound  =  312.3  +  (0.80)  (876.9)  =  1013  B.  t.  u., 
or  5065  B.  t.  u.  for  5  pounds. 

The  total  heat  for  x  =  1.00  is  312.3  +  876.9  =  1189.2  B.  t.  u.  per  pound, 
or  5946  B.  t.  u.  for  5  pounds. 

Therefore  the  heat  which  must  be  added  is  5946  -  5065  =  881  B.  t.  u. 
Byeq.  (49), 

the  volume  per  pound  =  (0.80)  (3.723)  +  (1  -  0.80)  (0.016)  =  2.99  cubic  feet. 

18.  Isoenergic  transformation.     If  three  pounds  of  a  mixture  of 
steam  and  water  at  a  pressure  of  120  pounds  absolute  and  a  quality  of 
0.90  undergoes  an  isoenergic  change  until  the  pressure  is  150  pounds 
absolute,  find  the  quality  and  the  volume  in  the  final  condition. 


76  THERMODYNAMICS 

Solution.     From  Table  3, 

For  120  pounds  For  150  pounds 

qi  =  312.3  g2  =  330.0 

Pi  =  794.2  P2  =  779.3 

s2  =  3.014 
Given  xi  =  0.90. 
By  eq.  (51),  ql  +  x1Pi  =  g2  + 


Therefore  x,  =  312.3  +(0.90)  (794.2)-  330.0  =  Q  ^ 

7  7  y.o 

By  eq.  (49),  #2  =  #2$2  +  (1  ~  a^V- 

Therefore  v2  =  (0.893)  (3.014)  +  (1  -  0.893)<r  per  pound 

=  2.70  cubic  feet  per  pound, 
or  for  3  pounds  the  volume  is  8.10  cubic  feet. 

19.  Isothermal  transformation.  Three  pounds  of  steam  and  water 
at  140  pounds  absolute  occupy  6  cubic  feet.  If  the  volume  is  in- 
creased at  constant  pressure  to  9  cubic  feet,  find  the  heat  added,  the 
external  work  done,  the  final  quality,  and  the  increase  in  entropy. 

Solution.  The  volume  per  pound  is  2  cubic  feet  in  state  1  and  3  cubic  feet  in 
state  2. 

From  Table  3,  for  140  pounds,  • 

p=  784.0,        Apu  =  83.4,         0V  =  1.0675,        si  =  3.220. 
By  eq..  (49),  vi  =  xis  +  (1  -  ZI)<T. 

Therefore  '          *  =  _=M_  =  0.6*0.  :, 


Similarly<  ^=3 

The  heat  added  is 

\2—  \l  =  x2r  —  xir  =  fa  —  xi)(p  +  Apu}, 
or 

(0.932  —  0.620)  (784  +  83.4)  =  245  +  26  =  271  B.  t.  u.  per  pound. 

Of  this  added  heat  26  B.  t.  u.  represents  the  external  work.     For  3  pounds  the  added 
heat  is  813  B.  t.  u.  and  the  external  work  is 

(3)  (778)  (26)  =  60,400  foot  pounds. 

The  increase  in  entropy  is  (x2  —  £i)0v  =  (0.932  —  0.620)  (1.0675)  =  0.333  per  pound, 
or  1.0  unit  of  entropy  for  3  pounds. 

20.  Adiabatic  transformation.  Five  pounds  of  steam  of  quality 
0.80  expand  adiabatically  from  130  pounds  to  30  pounds  absolute 
pressure.  Find  the  quality,  the  change  in  volume,  and  the  external 
work  done. 

Solution. 

For  130  pounds  For  30  pounds 

qi  =  318.6  g2  =  219.1 

Pi  =  789.0  P2  =  868.2 

0W1  =  0.5000    >  s2  =  13.74 

0ri  =  1.0808  0W,  =  0.3687 

«i  =  3.451  <f>V2  =  1.3305 

s2  =  13.74 


WATER   VAPOR  77 


Given  xi  =  0.80. 

By  eq.  (55),  for  an  adiabatic  expansion, 


Therefore  *a  =  0.6000  +  (0.80)  (1^808)  -  0.3687  =  ^ 

l.ooOo 

By  eq.  (56)  ,  the  work  per  pound  is 

ei  -  *2  =  —  (0i  +  *iPi  -  02  —  X2p2), 

^1 

or 

778  [318.6  +  (0.80)  (789)  -219.1  -  (0.75)  (868.2)]  =  778(79.5)  =62,000  ft.  lb., 
or  310,000  foot  pounds  for  5  pounds  of  mixture. 

By  eq.  (49),  t?i  =  (0.80)  (3.  450)  +  (1  -  0.80)<r, 

and  v2  =  (0.75)  (13.74)  +  (1  -  0.76)<r. 

The  increase  of  volume  per  pound  is  (neglecting  <r), 

(0.75)  (13.  74)  —  (0.80)  (3.45)  =  7.63  cubic  feet,  or  38.15  for  5  pounds. 

21.  Constant   volume  transformation.     A    closed  tank    containing 
10  pounds  of  dry  steam  at  130  pounds  absolute  is  cooled  until  90  per 
cent  of  the  steam  has  been  condensed.     What  is  the  final  tempera- 
ture and  how  much  heat  has  been  removed? 

Solution.  Since  the  change  takes  place  at  constant  volume,  no  external  work  is 
done,  and  the  heat  removed  is  that  which  is  required  to  reduce  the  intrinsic  energy 
from  state  1  to  state  2.  ^  . 

For  130  pounds 
0i  =  318.6 
Pi  =  789.0 
Si  =  3.451 

Given  x\  =1.00  and  x2  =  0.10.     Solving  for  1  pound, 
Byeq.  (49), 

vi  =  si  and  v-2  =  x.2s2  +  (1  —  £2)<r  =  0.10  s2  +  (0.90)  (0.016). 
For  constant  volume,  vi  =  t>2, 

hence  3.451  =  0.10  s2  +  0.014,         or  s.2  =  34.37. 

From  tables  similar  to  Table  3,  s2  =  34.37  is  found  to  be  the  specific  volume  of 
steam  at  198.9°  Fahrenheit  and  11.3  pounds  pressure. 

For  these  conditions,  g2  =  167.1,  p2  =  905.6. 

The  heat  removed  is  ei  —  e2,  where,  by  eq.  (51), 

6!  =  318.6  +  789.0  and  e2  =  167.1  +  (0.10)  (905.6); 
that  is,  850  B.t.u.  per  pound,  or  8500  B.t.u.  per  10  pounds. 

22.  Boiler  explosion.     How  much  energy  is  liberated  if  a  boiler 
containing  two  tons  of  water  at  353°  Fahrenheit  explodes  ?     Assume 
the   atmospheric  pressure  to   be  14.7  pounds.     Draw  the  0-<f>  dia- 


78 


THERMODYNAMICS 


gram.     What  volume  would  the   steam  thus   produced   occupy   if 
there  was  no  condensation? 


Solution. 

At  353°  F. 

at  14.7  pounds, 

,e 

353' 


p  =  140  pounds, 
qz  =  180.3. 


=  324.4 ; 


212' 


32' 


Hence  the  heat  energy 
liberated  per  pound  is 
qi  _  g2j  or  144  B.  t.  u. 

For  2  tons  this  is 
576,000  B.t.u.,  that  is 
448,000,000  foot  pounds. 

The  6-<t>  plot  is  shown 
to  scale  in  Figure  33, 
where  the  heat  liberated 
is  represented  by  the 
shaded  area. 

If  this  energy  is  used 
entirely  to  produce  re- 
evaporation  at  a  pressure 
of  14.7  pounds,  the  num- 
ber of  pounds  of  steam 
produced  is  found  as  fol- 
lows. From  Table  3,  r 
for  14.7  pounds  is  969.7. 
For  each  pound  of  water 
there  is  then  evaporated 
x  =  if |  =  0.149  pound. 
The  volume  occupied  by 
this  steam  is  0.149  s  or 
0.149  (26.78)  =  3.98  cubic 
feet.  The  steam  pro- 
duced by  the  explosion 
would  fill  4000  (3.98)  or 
15,920  cubic  feet  at  the 
atmospheric  pressure. 
Actually  the  steam  would 
in  part  be  formed  at 
^  higher  pressure,  there 
would  be  destructive  ex- 
pansion, and  the  values 
calculated  above  repre- 
sent not  actual  conditions  but  theoretical  conditions  which  are  useful  only  in  giving 
some  ideas  of  the  magnitude  of  the  energy  liberated. 

23.  Equivalent  evaporation  and  boiler  horse  power.  What  is  the 
equivalent  evaporation  from  and  at  212°  Fahrenheit  per  pound  of 
coal  in  a  boiler  where  19,000  pounds  of  feed  water  at  a  temperature 
of  70°  Fahrenheit  are  converted  into  steam  at  80.3  pounds  gauge  by 
the  combustion  of  one  ton  of  coal? 


*  FIG.  33 


WATER  VAPOK  79 

If  a  boiler  horse  power  is  defined  as  the  capacity  of  a  boiler  which 
evaporates  30  pounds  of  water  per  hour  from  a  temperature  of  100° 
Fahrenheit  against  a  pressure  of  70  pounds  gauge,  find  the  equiva- 
lent evaporation  from  and  at  212°  Fahrenheit,  which  is  equal  to  a 
boiler  horse  power.  If  the  above  evaporation  took  place  in  two 
hours,  what  is  the  boiler  horse  power  ? 

Solution.    From  Table  3, 

For  80.3  +  14.7  Ib.         For  70°  F.  For  70  +  14.7  Ib.  For  100°  F.        For  212°  F. 

q  =  294.6                   q  =  38.1                  q  =  286.4  q  =  68  q  =  180.3 

r  =  890.5                                                 r  =  896.7  r  =  969.7 

"i  Q  onn 

Per  pound  of  coal,  the  given  boiler  evaporates  —  2  -  =  9.5  pounds  of  water  from  70° 


Fahrenheit  to  steam  at  95   pounds  absolute.     The   B.  t.  u.    required  are  evidently 
294.6  +  890.5  —  38.1  =  1147.0  per  pound. 

Per  pound  of  water  from  and  at  212°  the  B.  t.  u.  required  are  969.7.     The  equiva- 

lent evaporation  of  the  given  boiler  is  evidently  9.5  -  —  =  11.25  pounds  of  water 
from  and  at  212°  per  pound  of  coal. 

Now  a  boiler  horse  power  is  equivalent  to  30(286.4  +896.7  -  68)  =30(1115.0)  B.  t.  u. 
per  hour.  That  is,  a  boiler  horse  power  is  equal  to  an  equivalent  evaporation  of 

30  1115-°  =  34.5  pounds  per  hour. 
969.7 

The  actual  boiler  had  an  equivalent  evaporation  per  hour  of  =522  11.25  =  11,250 

-I  -I     OKA  2 

pounds.     Hence,  a  boiler  horse  power  of  —t     —  =  325. 

34.5 

24.  Surface  condenser.  A  surface  condenser  receives  steam  at  a 
pressure  of  4  pounds  absolute.  Twenty  pounds  of  water  at  60°  Fahr- 
enheit flow  through  the  pipes  of  the  condenser  for  every  pound  of 
steam  that  is  condensed.  What  is  the  temperature  of  the  feed  water 
leaving  the  condenser  ?  The  condenser  water  flows  to  a  Worthing- 
ton  cooling  tower,  where  it  trickles  down  through  tiles  that  cause  it 
to  spread  over  a  large  surface.  If  five  per  cent  is  evaporated  and 
carried  away  by  a  current  of  air  from  a  fan  passing  through  the 
tower,  other  things  being  equal,  what  would  be  the  final  tempera- 
ture of  the  remaining  95  per  cent  ? 

Solution.  For  4  pounds  pressure,  q  =  121.0  and  r  =  1005,5,  or  \!  =  1126.5  B.  t.  u. 
For  60°  Fahrenheit  temperature,  q  =  28.1  =  X2. 

Let  qx  represent  the  heat  of  the  liquid  per  pound  for  the  final  common  temperature 
of  the  steam  and  condenser  water. 

Then  \x  -  qx  =  20  (^  -  X2), 

or  21  qx  —  Xi  +  20  X2, 

gj  =  1126.5  +  662  =8pA 

21 
Hence  the  temperature  is  112°  Fahrenheit. 


80  THERMODYNAMICS 

To  evaporate  5  per  cent  of  a  pound  of  water  at  112°  Fahrenheit  requires  0.05  r  B.  t.  u. 
For  112°  Fahrenheit,  r  is  1029.  Hence  it  requires  51.5  B.  t.  u. 

The  abstraction  of  51.5  B.  t.  u.  from  the  remaining  95  per  cent  of  the  pound  to  re- 
evaporate  the  5  per  cent  reduces  the  heat  of  the  liquid  for  the  water  to  [(0.95)  (80.  2) 
—  51.5]  B.t.  u.  for  each  95  per  cent  of  a  pound.  Hence  per  pound  of  remaining 

water  q  is  (°-95)(80-2)-  51-5  =  26  B.  t.  u.     The  corresponding  temperature  is  26  +  32, 

0.95 

or  58°  Fahrenheit.    (Actually  the  temperature  of  the  cooling  water  and  the  condensed 
steam  are  not  exactly  the  same.) 

25.  Jet  condenser.  A  jet  condenser,  which  injects  cooling  water 
into  the  steam  which  is  to  be  condensed,  receives  steam  at  4  pounds 
absolute.  The  water  of  the  jet  is  60°  Fahrenheit.  If  the  feed  is 
120°  Fahrenheit,  how  much  condensing  water  is  used  per  pound  of 
steam  condensed  ? 

Solution.  Using  the  values  of  X  from  Problem  24  and  q  for  120°  Fahrenheit  as 
88  B.  t.  u.,  if  we  represent  by  m  the  mass  of  the  condensing  water  used  per  pound  of 
steam  condensed,  then 

(1126.5  -  88)  =  m  (88  -  28.1), 


OT  =          .6  -88 

88-28.1  ' 

26.  Rankine  cycle.  The  operations  of  a  stearn  engine  are,  essen- 
tially, the  admission  of  steam,  its  expansion,  and  its  final  expulsion. 
The  quantity  of  the  working  fluid  in  the  engine  is  not  always  the 
same,  and  for  this  reason  the  p-v  plot  of  the  engine  does  not  repre- 
sent a  closed  cycle  in  the  sense  in  which  the  cycle  of  the  Carnot  en- 
gine of  page  22  is  a  closed  cycle.  Considering  the  boiler,  the  engine, 
and  the  condenser  as  a  unit,  the  steam  does  pass  through  a  closed 
cycle.  The  assumptions  of  the  theoretical  and  ideal  Rankine  cycle 
are  as  follows  :  (1)  Water  is  raised  from  the  temperature  at  which 
it  is  fed  into  the  boiler  to  the  temperature  and  pressure  at  which  it 
is  to  be  admitted  to  the  engine  cylinder.  (2)  When  the  admission 
valve  of  the  engine  opens,  an  isothermal  expansion  occurs  in  the 
boiler  and  engine  cylinder  while  the  piston  is  moved  forward  by  this 
expansion  until  c*ut-on°  occurs.  (3)  The  steam  then  expands  adia- 
batically  in  the  engine  cylinder  until  its  pressure  is  that  of  the  con- 
denser. (4)  It  is  then  expelled  by  the  back  stroke  of  the  piston 
and  condensed  at  the  constant  temperature  and  pressure  of  the  con- 
denser. This  condensed  water  is  then  fed  back  into  the  boiler. 

Draw  the  p-v  and  &-<]>  plots  of  such  a  Rankine  cycle  for  one  pound 
of  water  received  by  the  engine  and  expelled  at  a  condenser  pressure 
such  that  the  vacuum  gauge  reads  24.4  inches  of  mercury.  The 
boiler  pressure  is  80  pounds  gauge,  the  atmospheric  pressure  is 
15  pounds.  Calculate  the  efficiency. 


WATER   VAPOR 


81 


Solution.  The  p-v  plot  is  shown  in  Figure  35,  which  is  drawn  to  scale  to  emphasize 
the  volume  relations.  The  6-$  plot  is  also  drawn  to  scale  and  is  shown  in  Figure  34. 
The  work  done  may  most  easily  be  found  from  the  0-0  plot.  The  shaded  area  abed 
represents  the  work  done  in  heat  units.  The  efficiency  is  evidently 

area  abed  _  areas  (Oo5/+  beef—  Ooag  —  adeg} 
area  abceg  areas  (  Oobf  +  beef  —  Ooag) 


324 


142' 


32' 


O 


e 


FIG.  34 


Replacing  these  areas  by  their  heat  equivalents,  we  have 

_  gi  +  ri  —  q2  —  X2r2         _  ^ £2r2 

gi  +  **i  —  ga  gi  +  i"i  —  gs 

The  boiler  pressure  is  80  +  15  =  95  pounds  absolute.     Since  30  inches  of  mercury 
correspond  to  a  pressure   of   14.7  pounds,   24.4   inches  correspond  to   a  condenser 

vacuum  of  — —  24.4  =  12  pounds.     Hence  the  condenser  pressure  is  3  pounds  absolute. 


82 


THERMODYNAMICS 


95* 


From  Table  3, 
For  95  pounds  pressure 


=  0.4699 
=  1.1363 
=  4.644 


For  3  pounds  pressure 

q2  =  109.6 

r2  =  1012.2 

0W  =  0.2011 

0«a  =  1.6841 

S2  =  118.4 


4.6 


97.6cu.ft 


FIG.  35 


For  the  problem  under  solution  the  value  of  x2  must  be  found  by  using  equation 
(55),  the  condition  for  an  isentropic  expansion.     Thus 


or 


_  0.4699  +  1. 1363  -  0.2011      n  QQ/1 
*2  ~ L6841 °-834' 


18.3  cu.ft 


FIG.  36 
Then  substitution  in  the  equation  above  for  the  "  Rankine  efficiency  "  gives 

,  =  1 (0.834)  (1012.2)       =     _     ?       or  cent< 

294.6  +  890.5  -  109.6 

27.  Incomplete  expansion.  Because  of  the  large  piston  displace- 
ment required  for  a  complete  expansion  down  to  the  condenser 
pressure,  the  resulting  small  force  exerted  upon  the  piston  and  the 


WATER  VAPOR, 


83 


324° 


2289 


larger  amount  of  friction,  better  mechanical  efficiency  can  be  obtained 
by  sacrificing  some  theoretical  thermal  efficiency  and  stopping  the 
expansion  at  some  pressure  higher  than  the  condenser  back  pressure. 
Such  a  cycle  may  be  represented  as  in  Figure  36.  From  c  to  d  the 
steam  in  an  actual  engine  flows 
through  the  exhaust  valve  until 
the  pressure  has  fallen  to  the  con- 
denser back  pressure.  From  d  to 
e  the  remainder  of  the  steam  is 
expelled  by  the  back  stroke  of  the 
engine.  For  all  thermodynamic 
purposes  in  this  discussion  the 
effect  is  the  same  as  if  the  steam 
were  cooled  at  constant  volume 
from  c  to  d,  and  compressed  iso- 
thermally  from  d  to  e. 

If  the  initial  pressure  is  95 
pounds  and  the  condenser  back 
pressure  3  pounds  absolute  as  in 
Problem  26,  find  the  thermal  effi- 
ciency if  release  occurs  at  20 
pressure  absolute.  Draw  the  0-<f> 
plot.  What  proportion  of  the 
total  piston  displacement  is  the 
cut-off  in  this  case  ? 


FIG.  37 


Solution.  The  p-v  and  0-0  plots  are 
shown  to  scale  in  Figure  36  and  Figure  37.  The  efficiency  may  be  found  from  the  0-0 
plot  by  the  use  of  a  planimeter.  The  line  cd  represents  the  constant  volume  of  vCf 
The  efficiency  may  also  be  obtained  by  calculations  based  on  the  p-v  plot  as  follows. 
Denote  by  1  the  state  for  95  pounds  pressure,  by  2  that  for  20  pounds,  and  by  3  that 
for  3  pounds. 

_  areas  (nbmo  +  bcnm  —  dnoe} 
-  —  -  - 


JL  J.1C11 

where 

'i 
area  abmo 

=  j5ava  =  jOi«i  per  pound  of  entering  steam, 

area  bcnm 

=  J(<lb  +  Pi  —  qc  —  xcpc)  =  J(qi  +  pi  —  #2  - 

area  dnoe 

=  PdVd  =  Ps^c  =  Ps(.X2U2  +  0"). 

From 

Table  3, 

For  j 

Pi  =  95 

For^2  =  20                            For^ 

( 

71  =  294.6 

q2  =  196.4                              <a 

ri=890.5 

r2  =  959.4 

si  =  4.644 

s2  =  20.09 

0 

>Wi  =  0.4699 

M2  =  20.07 

t) 

^=  1.1363 

0W  =0.3362 

2 

Pi  =  808.8 

0^  =  1.3957 

Pa  =  885.1 

=  3 
qs  =  109.6 


84  THERMODYNAMICS 

For  an  adiabatic  expansion  to  a  pressure  of  20  pounds  and  quality  xz  we  have 


x,  =  0.4699+1.1363-0.8362  =  Q>m 

1.3957 

Hence  from  the  relations  established  from  Figure  36 

Work  during  admission  is  (95)  (144)  (4.644)  =  63,200  ft.  Ib.  per  Ib.  of  steam, 

Work  during  expansion  is  by  eq.  (36) 

778  [294.6+808.8-  196.4-  (0.909)  (885.1)]=  79,400  ft.  Ib. 
The  combined  work  of  admission  and  expansion  is  142,600  ft.  Ib. 
Work  during  expulsion  is  (3)  (144)  [0.909  (20.07)]  =  7800  ft.  Ib. 
Therefore  external  work  done  by  the  engine  is  134,800  ft.  Ib.  per  Ib.  of  steam. 

The  input  is  778(294.6  +  890.5  -  109.6)  =  832,000  ft.  Ib.  per  Ib.  of  steam. 


Hence  the  efficiency  is  =  16.1  per  cent. 

8320 

The  volume  at  release  per  pound  is  vc  =  (0.909)  (20.07)  =  18.3  cubic  feet. 
The  volume  at  cut-off  per  pound  is  vb  =  4.64. 

The  cut-off  is  then  -^-  =  25  per  cent. 
18.3 

28.  Efficiency  and  boiler  pressure.  Assuming  a  non-conducting 
engine  working  non-expansively  and  non-conducting,  draw  the  p-v 
and  #-$  plot  for  a  boiler  pressure  of  90  pounds  absolute  and  an 
atmospheric  pressure  of  14.7  pounds.  The  feed  water  is  60°  Fahren- 
heit. Calculate  the  efficiency  for 
pressures  of  50,  70,  90,  110,  arid  130 
pounds  absolute  and  plot  a  curve 
showing  the  relation  between  boiler 
pressure  efficiency  for  this  case. 

Solution.  The  efficiency  for  a  pressure  of  90 
pounds  is  calculated  as  an  illustration.  The  p-v 
and  0-0  plots  are  shown  in  Figures  38  and  39 


90* 


4.89  cu.ft. 


FIG.  38 


FIG.  39 


WATER   VAPOR 


85 


respectively.     Table  3  gives  for  90  pounds  g1=290.7,    ri  =  893.6,   and 
For  60°  Fahrenheit  qs  =  28.1. 

The  input  is  gx  +  n  -  q*  =  290.7  +  893.5  -  28.1  =  1156.1  B.t.u. 

The  output  is  most  conveniently  found  from  the  p-v  plot.    It  is  (pa—pd)Vb  — 


=  4. 


10 

Is 


5O 


60 


70 


80 


90 
Pounds 


FIG.  40 


100 


110 


120 


130 


Or  substituting  the  indicated  values,  the  output  is 

(144)  (90  -  14.7)4.886  =  52,900  foot  pounds  =  68  B.  t.  u. 


The  efficiency  is  then 


1156 


=  5.9  per  cent. 


The  curve  of  efficiency  and  boiler  pressure  is  shown  in  Figure  40. 


25 


20 


15 


10 


10  20 

Inches  of  Mercury 
FIG.  41 


86 


THEKMODYNAMICS 


29.  Condensing  engine.     Find  for  the  Rankine  cycle  the  theoreti- 
cal efficiency  of  an  engine  supplied  with  steam  at  100  pounds  abso- 
lute for  the  following  values  of  the  condenser  back  pressure :  1,  2,  3, 
4,  6,  10,  and  14.7  pounds  absolute.     Plot  efficiency  against  inches 
of  mercury  indicated  by  a  vacuum  gauge  from  an  atmospheric  press- 
ure of  14.7  pounds. 

Solution.    The  solution  of  this  problem  is  identical  with  that  of  Problem  26.     The 
plot  of  Figure  41  shows  the  variation  of  the  efficiency  with  the  condenser  vacuum. 

30.  Use  of  the  temperature  entropy  chart.     Assume  a  frictionless 
non-conducting  engine  working  on  a  cycle  of  incomplete  expansion 
as  in  Problem  27.     If  the  clearance  is  negligible,  the  cut-off  40  per 
cent  of  the  piston  displacement  of  5  cubic  feet,  the  admission  steam 
at  100  pounds  absolute,  and  the  condenser  back  pressure  3  pounds 
absolute,  draw  the  p-v  and  Q-$  plots. 


Solution.     One  pound  of  steam  at  100  pounds  pressure  has  a  volume  of  4.432  cubic 
feet. 


Therefore  at  cut-off  the  engine  cylinder  contains  — —  =  0.451  pounds. 

100  4.432 


328° 


259° 


Since  the  0-0  chart  and  the  tables  for 
steam  given  on  page  130  are  both  per  pound 
of  steam,  it  is  most  convenient  to  reduce 
the  engine  under  consideration  to  one  con- 
taining a  pound  of  steam  (or  mixture  if 
the  steam  is  wet).  Thus,  if  the  cylinder 
volume  is  imagined  to  be  increased  in  the 
1  5.00 


ratio 


that  is,  to 


=  11.1  cubic 


32 


0.451  0.451 

feet,  there  will  be  one  pound  of  steam  in 
the  cylinder  at  cut-off. 

From  the  0-(f>  chart  of  Figure  42  it  is 
found  that  the  entropy  of  one  pound  of 
dry  steam  at  100  pounds  is  1.602.  Where 
the  line  of  constant  entropy  of  this  value 
intersects  the  line  of  constant  volume  for 
11  cubic  feet  gives  the  point  c  on  the  6-(f> 
plot  of  Figure  43,  corresponding  to  the 
point  c,  which  may  now  be  plotted  on  the 
p-v  plot  of  Figure  44.  If  intervening 
points  on  the  p-v  plot  of  the  expansion  are 
desired,  they  may  be  found  by  noting  the 
intersections  of  the  line  of  constant  en- 
tropy, equal  to  1.602,  with  various  lines  of 
constant  volume  and  then  finding  the 
pressures  corresponding  to  those  intersec- 
tions. The  constant  volume  line  of  value 

11  cubic  feet  is  followed  until  it  intersects  the  constant  pressure  line  for  3  pounds. 

This  gives  the  point  d  on  the  0-<£  plot. 


FIG.  43 


Sj 
'088 

$175 

•'oae 

T  140 
130 


[110 
[  100 
N£ 
80 


8S 


oas 


oss 


*081 


I 


<ae.i 


08.1 


Q. 

'OS! 


001 


100 


1.42        1.44        1.46        1.48        1.50        1.52        1.54        1.56        1.58        I 


1.62         1.64        1.66         1.68 


1.74        1.76         1.78 


oos 


on 

0£l 


Oil 

OOl 


08 


' 


oa 

06 1 


06- 


0! 

:; 


3 

e| 

f- 


8T.r~ 


WATEK  VAPOR- 


87 


The  efficiency  may  now  be  found  after  the  manner  of  Problem  27.  The  pressure 
volume  plot  for  this  engine  with  its  piston  displacement  of  5  cubic  feet  is  shown  dotted 
in  Figure  44. 


100" 


35* 


\ 


\ 


\   \, 

c 

d 

V 

-     -  -                    1 
1 
1 
1 

I 

5                                           11.1  CU.ft. 

FIG.  44 


31.  Boulvin  dia- 
gram. A  simple 
method  of  transform- 
ing graphically  a  steam 
cycle  expressed  in 
terms  of  any  one  of  the 
following  sets  of  co- 
ordinates into  terms  of 
any  other  set  is  known 
from  its  proposer  as 
the  Boulvin  diagram. 
The  coordinates  are  p 
and  v,  p  and  0,  0  and 
<£,  and  (£  and  v.  If 
four  quadrants  are  drawn,  as  in  Figure  45,  the  method  may  be  shown 
by  changing  a  point  &m  expressed  in  terms  of  p  and  v  into  each  of  the 

other  sets  of  coor- 
dinates. Thus, 
in  quadrant  II, 
draw  the  curve 
oc  for  the  press- 
ure-temperature 
relations  for  the 
boiling  point  of 
water.  To  any 
point  km  in  the 
pressure -volume 
quadrant  there 
is  possible  but 
one  temperature, 
namely,  that  cor- 
responding to  its 
pressure  or  to 
the  temperature 
of  ^  on  the  tem- 
per a  tu  re-press- 
ure curve  of 
FIG.  45  quadrant  II. 


88  THERMODYNAMICS 

Now  draw  in  quadrant  I  the  0-(f>  curve  ab  for  this  temperature.  The 
line  of  constant  temperature  a'b'  in  quadrant  IV  may  then  be  drawn 
as  described  in  Section  51.  The  projection  of  Jcm  on  the  curve  a'b'  at 
&iv  then  gives  the  entropy  corresponding  to  the  pressure  (that  is, 
to  the  temperature)  and  volume  of  the  point  &m.  The  point  ^  on 
the  0-(p>  plot  is  then  found  by  projecting  Jciy  on  ab. 

Figure  46  shows  the  indicator  card  of  an  actual  engine.  The 
piston  displacement  was  1.037  cubic  feet.  Compression  occurred 
at  3.8  per  cent  stroke.  The  amount  of  steam  supplied  per  stroke  to 
the  end  of  the  cylinder  for  which  this  card  was  obtained  was  0.0781 


50*SPRING 


FIG.  46 


pound.  The  clearance  was  9.4  per  cent  of  the  piston  displacement. 
The  receiver  back  pressure  was  14.7  pounds.  Draw  the  #-</>  plot 
for  the  expansion  and  discuss  the  heat  transfer  indicated. 


Solution.  Assume  the  steam  dry  at  compression.  This  is  the  ordinary  assumption 
of  steam  engineering  practice.  All  the  water  in  the  cylinder  at  release  is  supposed  to 
reevaporate  during  the  back  stroke  of  the  piston  and  be  expelled,  except  for  that  steam 
which  remains  in  the  clearance  spaces.1  The  steam  at  compression  then  occupies 
(9.4  +  3.8)  per  cent  of  1.037  cubic  foot  at  a  pressure  of  14.7  pounds.  For  this  pressure 
the  specific  weight  or  density  is  found  from  Table  3  to  be  0.0374.  Hence  the  clearance 
steam  was  T\^(  1.037)  (0.0374)=  0.0052  pound.  At  cut-off  there  was  then  present  in 
the  cylinder  0.0781  +  0.0052  =  0.0833  pound  of  steam. 

Since  the  6-$  diagram  is  most  conveniently  plotted  for  unit  weight  of  steam,  we 


1  For  further  discussion  of  the  quality  at  compression  see  Reeve,  "Thermodynamics  of 
Heat  Engines,"  p.  214  (Macmillan) . 


WATEK   VAPOR 


89 


FIG.  47 


90 


THERMODYNAMICS 


follow  the  method  of  Problem  30  and  plot  in  the  p-v  quadrant  of  the  Boulvin  diagram 
the  changes  in  the  pressure  and  volume  per  pound  of  mixture.  That  is,  all  the  volumes 
of  the  actual  problem  are  increased  in  the  ratio  of  1  to  0.0833.  The  p-v  plot  of  quadrant 
III  of  Figure  47  shows  the  expansion  line  of  the  indicator  card  of  Figure  46  so  modified 


and  plotted.     The  piston  displacement  has  been  increased  to 


1.037 


0.0833 
Similarly,  the  clearance  has  become  9.4  per  cent  of  12.45,  or  1.17  cubic  feet, 


=  12.45  cubic  feet. 


FIG.  48 

The  6-<}>  plot  of  the  expansion  has  been  constructed  by  the  method  outlined  above. 
The  construction  lines  for  two  points  which  have  been  reproduced  in  the  plate  indicate 
sufficiently  the  method. 

Discussion.  The  steam  supplied  to  this  engine  was  found  by  means  of  a  throttling 
calorimeter  to  have  a  quality  of  99  per  cent  when  it  entered  the  cylinder.  Due  to  con- 
densation the  quality  had  fallen  to  47  per  cent  at  cut-off.  Some  further  condensation 
took  place  during  the  earlier  part  of  the  expansion.  For  the  latter  and  greater  part  of 
the  expansion,  however,  heat  was  returned  to  the  steam  from  the  cylinder  walls.1  The 
quality  therefore  rose  to  72  per  cent  at  release. 

1  That  this  heat  is  at  a  lower  availability  as  a  result  of  the  lower  temperature  is  evident 
from  Problem  50.  The  present  problem  serves  as  an  introduction  to  the  discussion  of  the 
gain  due  to  superheating  which  is  made  in  Section  62  of  page  98. 


WATER   VAPOR  91 

32.  Hyperbolic  expansion  line.  Because  the  actual  adiabatic  line 
for  a  steam  transformation  is  not  to  be  drawn  by  any  simple  graphi- 
cal method,  it  is  the  custom  of  steam  engineers  to  judge  of  the 
actual  operation  of  a  given  engine  by  drawing  through  the  point  of 
cut-off  on  the  p-v  plot  (indicator  card)  of  the  engine  a  hyperbola. 
It  is  the  object  of  this  problem  to  compare  the  hyperbolic  line  with 
the  true  isentropic.  By  using  the  Boulvin  diagram  find  the  p-v  plot 
of  an  isentropic  line  which  has  been  assumed  in  the  6-<f>  quadrant. 
Then  through  the  point  of  this  adiabatic  in  the  p-v  quadrant  which 
corresponds  to  the  highest  assumed  pressure  draw  a  rectangular 
hyperbola. 

Solution.  The  construction  of  the  adiabatic  ab  is  evident  from  Figure  48.  To 
draw  the  hyperbola  proceed  as  follows.  Draw  amm  perpendicular  to  Ov  and  amn  per- 
pendicular to  Op.  Through  some  point  vc  on  Ov  draw  vck  perpendicular  to  Ov. 
At  k  the  intersection  of  vck  and  amn  draw  kO  to  the  origin.  Through  j,  the  intersection 
of  kO  and  c^w,  drawp^'  perpendicular  to  Op  to  meet  vck  in  c,  which  is  a  second  point 
on  the  hyperbola.  In  a  similar  manner  other  points  may  be  obtained. 


CHAPTER  IV. 

SUPERHEATED   STEAM 

55.  Three  Molecular  States.     It  is  a  well-known  fact  of  physics 
that  there  are  three  molecular  conditions  in  which  substances  may 
exist.     These  are  commonly  stated  as  solid,  liquid,  and   aeriform. 
The  aeriform   bodies  are   commonly  subdivided  into    two  classes; 
namely,  those  which  obey  Boyle's  Law  at  ordinary  temperatures  and 
pressures  and  are  called  gases,  and  those  which  at  ordinary  pressures 
and  temperatures  can  exist  in  both  a  liquid  and  an  aeriform  state 
and  are  called  vapors.       The  distinction  between   these  molecular 
conditions  is  not  one  depending  upon  the  substance  considered,  but 
is  one  dependent  wholly  upon  its  physical  conditions  of  pressure  and 
temperature. 

Thus  it  is  known  that  the  so-called  "  permanent  gases  "  like  hydro- 
gen and  oxygen  have  been  liquefied  and  even  solidified  at  extremely 
low  temperatures  and  high  pressures.  For  any  gas  there  is  a  certain 
critical  temperature  above  which  it  cannot  be  liquefied.  And  at 
this  critical  temperature  there  is  a  certain  minimum  value  of  the 
pressure  which  must  be  applied  to  produce  liquefaction.  In  the 
development  of  this  portion  of  physics  the  term  "  permanent  gases  " 
was  introduced  and  applied  to  those  substances  in  an  aeriform  con- 
dition which  resisted  the  efforts  that  had  been  made  to  liquefy  them. 
The  term  to-day,  since  the  work  of  Wroblewski  and  Olszewski, 
Pictet,  Cailletet,  Dewar,  and  Liride,  is  evidently  a  misnomer. 

The  same  substance  can  exist  in  any  one  of  the  three  principal  mo- 
lecular states.  The  commonest  example  of  this  is,  of  course,  water, 
which  at  ordinary  temperatures  may  coexist  in  all  three  states. 
For  the  ordinary  temperatures  of  our  atmosphere,  however,  the  aeri- 
form state  which  water  assumes  must  be  classed  as  a  vapor  state. 
It  seems  reasonable  to  consider  that  at  higher  temperatures  it  would 
more  nearly  approximate  a  true  gaseous  state. 

56.  Superheated  Steam.     Water  vapor,  as  has  already  been  noted 
in  previous  discussions,  obeys  the  ordinary  vapor  laws  by  having  the 
temperature  and  pressure  of  the  liquid  from  which  it  is  produced  and 
with  which  it  is  in  contact.     But  when  removed  from  contact  with 

92 


SUPERHEATED    STEAM 


93 


its  liquid  (whether  actually  or  essentially  as  described  in  Section  44), 
its  temperature  may  be  raised  above  its  former  value.  The  vapor 
or  steam  is  then  said  to  be  superheated.  According  to  that  portion 
of  the  molecular  theory  stated  above  it  is  to  be  expected,  then,  that 
the  pressure,  specific  volume,  and  temperature  of  superheated 
steam  would  tend  to  follow  an  equation  somewhat  of  the  form  of  the 
characteristic  equation  of  a  perfect  gas. 

In  accordance  with  this  supposition  two  very  satisfactory  formulas 
have  been  proposed.  They  are  both  of  the  form  pv  =  R6,  but  con- 
tain corrective  terms.  The  simplest,  and  for  ordinary  purposes  one 
of  sufficient  agreement  with  the  actual  experimental  results,  is  that 
suggested  by  Tumlirz  and  based  upon  the  experiments  of  Battelli. 
It  will  usually  be  referred  to  as  the  "equation  of  Tumlirz."  Ex- 
pressed in  the  units  of  the  F.  P.  S.  system  it  is 

;w  =  85.85  0-0.256;?,      .     .     .     .\..    (57) 

where  0  is  the  absolute  temperature  in  Fahrenheit  degrees  and  p 
and  v  are  specific  pressure  and  volume. 

The  equation  proposed  by  Knoblausch,  Linde,  and  Klebe  l  is  more 
satisfactory  in  its  agreement,  but  the  accuracy  is  attained  at  the 
expense  of  a  tedious  calculation.  It  is  expressed  in  the  same 
symbols  as  equation  (57)  as  follows : 

pv  =  85.85  0- XI  +  0.00000976p)(^^000 

57.  Specific  Heat  of  Superheated  Steam.  The  specific  heat  of  super- 
heated steam  has  been  determined  for  constant  pressure  conditions 
by  Knoblausch  and  Jakob.  The  following  table  of  their  values  is 
only  partial :  2  -, 


-0.0833 


).    (58) 


Pressure  #  per  sq.  in. 

p  =  85.3 

p  =  113.8 

p  =  142.2 

p  =  170.6 

p  =  199.1 

Temperature  °  F. 

t,  =  316 

t,  =  336 

t0  =  354 

t0  =  368 

t0  =  381 

h  =  392 

0.530 

0.560 

0.597 

0.635 

0.677 

h  =  482 

0.514 

0.532  ( 

0.552 

0.570 

0.588 

ti  =  572 

0.505 

0.517 

0.530 

0.541 

0.550 

The  table  gives  the  average  value  of  the  specific  heat  (for  the  press- 
ure indicated  in  the  upper  row)  which  should  be  used  in  calculating 

1  These  equations  expressed  in  F.  P.  S.  units  are  taken  from  the  "  Steam  and  Entropy 
Tables  "  of  Professor  Peabody. 

2  A  complete  table  in  F.  P.  S.  units  is  given  in  Professor  Peabody's  "Tables." 


94  THERMODYNAMICS 

the  heat  required  to  superheat  from  the  saturation  temperature  (cor- 
responding to  this  pressure  and  given  in  the  second  row)  to  the 
temperature  given  in  the  left-hand  column.  Illustration  of  the  use 
of  this  table  will  be  given  in  Problem  35. 

For  all  engineering  purposes  steam  is  superheated  at  constant 
pressure.  The  additional  heat  required  to  raise  one  pound  of  dry 
steam  from  the  condition  of  saturated  vapor  at  a  pressure  p^  and 
a  temperature  tQ  (or  #0  absolute)  to  a  temperature  t  (or  0*)  will  be 
denoted  by  the  symbol  h.  That  is, 

A  =  *p(0-0o)  =  <*(*-*<>)>  .....     (59) 

where  cp  represents  the  specific  heat  at  constant  pressure  as  found 
from  the  above  table.  The  total  heat  contents  of  a  pound  of  super- 
heated steam  at  a  temperature  t  is  then 

*=q*  +  rQ  +  k       ......     (60) 

where  rQ  and  qQ  are  the  values  for  the  saturation  temperature  1  0 
which  corresponds  to  the  pressure  p  (equal  to  jp0,  since  the  super- 
heating is  at  constant  pressure). 

For  many  engineering  purposes  the  value  of  cp  may  be  taken  as 
0.48  with  sufficient  accuracy. 

58.  Entropy  of  Superheated  Steam.  The  increase  in  entropy  due 
to  superheating,  expressed  by  the  symbol  <£5,  is  sufficiently  well  ob- 
tained for  most  purposes  l  by  using  the  average  value  of  the  absolute 
temperature  at  which  the  superheating  occurs  and  the  specific  heat  cp 
as  0.48.  Thus, 


2  2 

The  total  entropy  <£  of  a  pound  of  superheated  steam  as  measured 
from  the  assumed  zero  of  entropy  at  32°  Fahrenheit  may  then  be 

written  *-*.+  *.  +  *.  .......     (62) 

59.  Temperature-Entropy  Chart  for  Superheated  Steam.  The  O-fy 
chart  for  steam  may  now  be  extended  into  the  superheated  region 
to  the  right  of  the  curve  of  constant  steam  weight.  The  increase  of 
entropy  due  to  superheating  at  constant  pressure  is  found  by  equa- 
tion (61),  and  lines  of  constant  pressure  p^  p%,  and  so  on,  are  plotted 

1  Stodola,  in  the  2d  edition  of  "The  Steam  Turbine,"  uses  0.48.  Cross  interpolation  in 
the  table  given  on  page  93  is  usually  unsatisfactory  and  not  as  accurate  as  a  graphical 
process.  The  tables  of  Marks  and  Davis  give  values  of  u,  X,  and  0  for  intervals  of  10°  F. 
for  steam  superheated  at  all  the  ordinary  pressures  of  engineering. 


SUPERHEATED   STEAM 


95 


as  in  Figure  49.  These  lines,  representing  the  increase  in  entropy 
at  constant  pressure,  are  essentially  straight  for  small  ranges  of  tem- 
perature. For  large  ranges  and  for  high  temperatures  of  superheat, 
they  should  be  plotted,  using  values  of  the  entropy  as  determined 
more  exactly  by  use  of  the  specific  heat  values  obtained  by  Knob- 
lausch  and  Jakob. 


FIG.  49 

Constant  volume  lines  may  also  be  extended  into  the  superheated 
region  by  calculating  for  any  assumed  pressure  and  volume  the  cor- 
responding temperature  0  in  accordance  with  either  equation  (57) 
or  (58).  The  intersection  of  the  line  for  constant  pressure  for  the 
assumed  value  and  the  line  of  constant  temperature  for  the  calculated 
value  0  gives  a  point  on  the  desired  constant  volume  line.  The  con- 
stant volume  line  might  be  drawn  by  a  method  similar  to  that  em- 
ployed in  constructing  the  constant  pressure  line  and  using  the 
increase  in  entropy  for  this  condition  calculated  in  a  similar  manner, 
were  it  not  that  the  value  of  cv  as  0.346  is  not  reliable.  Constant 
pressure  and  constant  volume  lines  are  shown  in  Figure  49. 

In  a  temperature-entropy  chart  constructed  as  above  it  is  evident 
from  Figure  49  that  both  the  lines  of  constant  pressure  and  the 


96  THEEMODYNAMICS 

lines  of  constant  volume  are  discontinuous  at  the  curve  of  constant 
steam  weight,  since  they  are  different  functions  of  the  temperature 
and  the  entropy  in  the  wet  steam  and  superheated  steam  regions. 
In  the  wet  steam  region  the  lines  of  constant  pressure  coincide  with 
the  lines  of  constant  temperature.  The  chart  is  therefore  equally 
well  a  #-<£  or  a  p-$  plot.  In  the  superheated  region  it  is  merely 
a  0-<f>  plot.  Because  of  the  fact  that  steam  is  usually  heated  at 
constant  pressure  a  p-<t>  plot  is  sometimes  convenient  in  this  re- 
gion. A  chart  constructed  in  this  way  is  shown  in  Figure  42  on 
page  86.  1  To  indicate  temperatures  it  is  necessary  to  draw  "quality  " 
lines  through  those  points  representing  steam  of  the  same  degree  of 
superheat.  Such  lines  would  be  everywhere  equally  distant  from 
the  curve  of  constant  steam  weight  if  the  specific  heat  of  superheated 
steam  were  constant.  In  the  chart  shown  these  lines  are  drawn  for 
intervals  of  superheat  equal  to  20  degrees.  The  scale  of  tempera- 
tures given  on  the  left-hand  side  obviously  refers  only  to  points 
within  the  wet  steam  region  or  on  the  dry  steam  line.  The  unequal 
scale  of  pressures  on  the  right  refers  to  any  point  of  the  plot.  Such 
a  plot  is  frequently  spoken  of  as  a  temperature-entropy  chart,  but  is 
one  only  for  the  region  of  wet  steam  and  more  rigorously  should  be 
called  a  pressure-entropy  chart  as  stated  above. 

60.  Intrinsic  Energy  of  Superheated  Steam.  Let  pl  be  the  press- 
ure upon  a  pound  of  superheated  steam,  t^  (or  6^  absolute)  be  its 
temperature,  and  v1  its  specific  volume  as  calculated  by  either  equa- 
tion (57)  or  (58).  This  superheated  steam  may  be  considered  as 
derived  at  constant  pressure  pQ  (equal  to  p^)  from  a  pound  of  dry 
saturated  steam  at  a  saturation  temperature  t0  (or  00)  corresponding 
to  this  pressure  pQ.  During  the  formation  of  this  pound  of  dry 
steam  a  quantity  of  intrinsic  energy  has  been  added  to  it,  which 
measured  from  water  at  32°  Fahrenheit  has  been  expressed  in  equa- 

tion (50)  as  e0  =  —  (<70+/90),  where  the  subscript  0  indicates  the  value 
A. 

of  any  magnitude  to  which  it  is  attached  corresponding  to  the  satu- 
ration temperature  tQ.  Since  p0  =  rQ  —  -ApQuQ  by  equation  (45)  and 
since  w0=s0  —  cr  by  equation  (44),  it  follows  that  the  intrinsic  energy 
of  the  pound  of  dry  saturated  steam  may  be  written  as 

-<)  .....     (0 


1  The  entropy  diagram  constructed  in  connection  with  Problem  50  on  page  120  is  drawn 
both  as  ap-(f>  and  as  a  0-0  plot  to  illustrate  this  difference. 


SUPERHEATED   STEAM 


97 


During  superheating,  at  the  pressure  pv  to  the  temperature  ^  the 
volume  has  increased  to  a  value  vv  and  there  has  been  done  an 
amount  p^v1  —  s0)  of  external  work.  Since  the  total  heat  added 
has  been  A,  the  increase  el  —  e0  in  intrinsic  energy  due  to  super- 
heating is  -j^ 

The  total  intrinsic  energy  el  possessed  by  a  pound  of  superheated 
steam  in  the  condition  indicated  by  the  subscript  1  is  then 


61.  Isentropic  Transformation.  From  the  6-<j>  plot  of  Figure  50 
it  is  evident  that  an  isentropic  or  reversible  adiabatic  expansion  of 
superheated  steam  may  be  considered  in  two  parts ;  namely,  an  ex- 


0 


Po=R 


m 


pansion  entirely  within  the 
superheated  region  until 
the  state  k  represented  by 
the  intersection  of  the  isen- 
tropic line  mn  and  the 
curve  of  constant  steam 
weight,  and  a  further  ex- 
pansion within  the  region 
of  wet  steam.  Dur- 
ing the  first  part  of  the 
'expansion  the  quality, 
which  may  now  be  taken 
to  mean  degrees  of  super- 
heat, that  is  t  — 10,  de- 
creases, and  during  the 
second  part  of  the  expan- 
sion the  quality  as  repre- 
sented by  x  decreases. 

If  the  expansion  is 
wholly  in  the  superheated  region,  the  quality  in  the  second  state 
expressed  in  degrees  superheat  may  most  easily  be  found  from  a  6-<f> 
chart.  It  may,  however,  be  found  analytically  by  the  method  il- 
lustrated in  Problem  38. 

If  the  expansion,  as  is  more  usual,  extends  into  the  region  of  wet 
steam,  the  quality  may  be  found  for  the  second  state  by  the  follow- 
ing relation  for  constant  entropy  : 


FIG. 50 


98  THERMODYNAMICS 

62.  Superheating  in  Engineering  Practice.  The  occasion  for  the 
use  of  superheated  steam  is  evident  in  part  from  Problem  31  on 
page  87.  The  steam  supplied  to  the  engine  there  described  had  a 
quality  of  over  99  per  cent  as  it  entered  the  engine.  Due  to  the 
fact  that  heat  was  lost  to  the  walls  of  the  engine  cylinder  its  quality 
had  dropped  to  47  per  cent  at  cut-off.  During  a  portion  of  the 
expansion  the  transfer  of  heat  to  the  cylinder  walls  continued,  but 
during  the  latter  part  of  the  expansion  some  of  this  heat  was  re- 
turned to  the  mixture,  reevaporating  some  of  the  53  per  cent  origi- 
nally condensed  during  admission.  During  exhaust  still  further 
heat  was  abstracted  from  the  cylinder  walls  to  reevaporate  (at  the 
condenser  back  pressure)  the  moisture  left  in  the  cylinder.  As  a 
result  the  temperature  of  the  cylinder  rises  with  the  admission  of 
steam  and  falls  with  its  exhaust. 

Consider  for  a  moment  that  the  entering  steam  is  dry,  if  the 
quality  at  cut-off  is  xv  then  the  intrinsic  energy  of  the  mixture  has 
been  reduced  from  q1  +  p1  to  qi  +  xtfi*  The  difference  (1  —  x1)p1 
then  represents  (neglecting  all  further  changes  due  to  heat  transfers 
to  the  cylinder)  a  decrease  in  energy  which  should  be  available  for 
the  expansion.  During  the  expansion  still  further  heat  energy  is 
lost  to  the  cylinder,  and  later  returned  at  a  lower  availability.  Of 
the  heat  lost  to  the  walls  during  admission,  part  is  returned  at  lower 
temperatures  during  expansion,  and  the  rest  in  reevap'oration  during 
exhaust.  All  of  this  energy  of  reevaporation  is  of  course  rejected, 
to  the  condenser. 

It  is  evident,  since  a  quantity  of  heat  q2  +  r2,  where  the  subscript 
2  represents  condenser  conditions,  is  always  to  be  rejected  to  the 
condenser,  that  as  much  as  possible  of  the  difference  q^  +  r^  —  q2  —  r% 
between  the  total  heat  supplied  and  rejected,  must  be  converted  into 
work.  To  accomplish  this  the  mean  temperature  of  the  cylinder 
must  be  more  nearly  that  of  the  entering  steam.  To  prevent  the 
loss  at  admission  the  steam  may  be  supplied  superheated  by  a  num- 
ber of  degrees  such  that  at  cut-off  the  steam  in  the  cylinder  is  dry. 
To  prevent  the  loss  during  expansion  the  entering  steam  may  be 
still  further  superheated  so  as  to  be  dry  at  release.  Then  no  heat 
will  be  abstracted  from  the  cylinder  walls  to  reevaporate  at  the  con- 
denser back  pressure.  For  any  given  engine  it  is  of  course  a  matter 
of  experiment  to  determine  how  much  superheat  is  required. 

In  general,  it  may  be  said,  superheated  steam  reduces  the  amount 
of  heat  interchanged  between  the  steam  and  the  cylinder  walls,  and 
thereby  reduces  the  weight  of  steam  required  per  stroke  of  the 


SUPERHEATED   STEAM  99 

engine  for  the  same  work  output.  Quoting  from  Ripper:  1  "Super- 
heating may  be  looked  upon  not  as  a  means  of  obtaining  a  thermal 
efficiency  with  the  engine  in  any  way  proportional  to  the  tempera- 
ture used  in  superheat,  but  as  a  device  for  realizing,  or  at  least  ap- 
proaching, the  full  thermal  efficiency  of  the  saturated  steam  between 
the  ranges  of  pressure  used  by  the  engine." 

In  the  case  of  the  steam  turbine  much  of  the  advantage  of  dry 
steam  is  the  consequent  small  friction  in  the  turbine  nozzle  and  pas- 
sages. The  presence  of  water  materially  increases  this  friction. 
Experiments  by  Stodola2  show  that  the  friction  increases  as  the 
density  of  the  steam  increases.  Thus,  for  example,  interpreting  some 
experiments  performed  by  Lewicki,  Stodola  says,  the  work  of  fric- 
tion of  a  wheel  in  saturated  steam  with  equal  specific  weight  (i.e. 
density  7),  equal  size  of  wheel,  and  equal  velocity  is  1.3  times  that 
in  the  air.  The  work  of  friction  for  a  wheel  in  superheated  steam 
at  atmospheric  pressure  and  572°  Fahrenheit  is  the  same  as  that  for 
air. 

63.    Resume  of  Equations  for  Superheated  Steam. 

Characteristic  equations. 

pv  =  85.85  0  -  0.256  p.     ...     .     .     .     ,    V    .  ,  .-   ..-    V    .     (57) 

pv  =  85.85  d  -  p  (1  +  Q.QOOQ0976ff)f15Q^^OQO  -  0.0833\       (58) 

\  v6  .  / 

Total  heat  contents. 

x=?o  +  ro  +  ^  =  2o  +  r0+^(<9-00).      .....    j     •     (60) 

Total  intrinsic  energy. 

«i=3*i-pO'i-<0  ...........   ,•     •     (63) 

Total  entropy. 


(62) 


1  Ripper,  "  Steam  Engine,  Theory  and  Practice,"  4th  ed.,  p.  156  (Macmillau) . 

2  Lowenstein's  translation,  Stodola,  "  The  Steam  Turbine,"  2d  ed.,  pp.  135-140. 


100  THERMODYNAMICS 


PROBLEMS  AND   SOLUTIONS:    SUPERHEATED    STEAM 

33.  Specific  volume.     Find  from  equations  (57)  and  (58)  the  vol- 
ume occupied  by  two  pounds  of  steam  superheated  75°  Fahrenheit 
at  a  pressure  of  135  pounds  absolute. 

Solution.     From  Table  3,  for  p  =  135  we  have  to  =  350D  F. 

The  absolute  temperature  of  the  superheated  steam  is  d  —  460  +  350  +  75  =  885. 
In  eq.  (58),  substitution  of  this  value  of  0  gives  for  the  corrective  term 
150,300,000  _  Q  0833\  _  Q  134 
0*  °) 

Hence     v  =  (85-85)(885)  _  [l  +  0.00000976(135) (144)] (0.134), 
(135) (144) 

or  v  =  3.91  —  0.16  =  3.75  cubic  feet. 

Or  the  volume  of  two  pounds  of  steam  is  7.50  cubic  feet. 

By  eq.  (57),        v  =  (85-85)(885)  _  0.256  =  3.91  -  0.26  =  3.65  cubic  feet. 
(135) (144) 

The  volume  for  two  pounds  is  then  7.30  cubic  feet. 

The  per  cent  of  difference  between  the  values  calculated  by  equation  (57)  and  (58) 

3.75-3.65  ,f 

is  then  —  100  =  2.7  per  cent. 

3.75 

34.  Temperature  of  superheat.     At  130.3  pounds  gauge  3  pounds 
of  steam  occupy  10.5  cubic  feet.     Find  the  degrees  of  superheat. 

Solution.  The  volume  per  pound  is  3.5  cubic  feet.  The  pressure  is  145  pounds  abso- 
lute. The  equation  of  Knoblauch,  Linde,  and  Klebe  (58)  is  capable  of  solution  only  by 
means  of  tedious  approximations.  The  equation  of  Turnlirz  (57)  is  here  used. 

,  =  P1  +  0.266J,  =  (145)(144)(8.6+0.256)  =  91go  Fahrenheit  absolute. 
85.85  85.85 

From  Table  3,  the  saturation  temperature  corresponding  to  a  pressure  of  145  pounds 
is  355.8°  F.  or  815°  F.  absolute.  The  superheat  is  therefore  913  —  815  =  98°  F. 

If  greater  accuracy  is  desired  and  justified  by  .the  precision  with  which  p  and  v  are 
known,  this  value  of  6  =  913  may  be  substituted  in  the  corrective  term  of  equation  (58) 
and  then  that  equation  solved  for  0,  considering  the  corrective  term  to  be  constant. 
Such  a  substitution  gives  0  =  885  and  70  degrees  of  superheat. 

35.  Specific  heat  of  superheated  steam.     Find  the  heat  required  to 
superheat  2  pounds  of  steam  75  degrees  at  135  pounds  absolute. 

Solution.     For  135  pounds  the  saturation  temperature  is  350°  Fahrenheit. 
From  Table  4,  the  specific  heat  is  found  to  be  approximately  0.558,  by  cross  inter- 
polation between  the  four  values. 


pounds 

113.8 

142.2 

392°  F. 

0.560 

0.597 

482°  F. 

0.532 

0.552 

SUPERHEATED   STEAM  101 

The  heat  required  per  pound  is  then  (0.558)  (75)=  41.8  or  42  B.  t.u.  For  two 
pounds  it  would  require  84  B.  t.  u. 

36.  Total  heat.     Find  the  total  heat  contents  of  two  pounds  of 
steam   superheated  75°  Fahrenheit    at   a   pressure    of   135    pounds 
absolute.     How  much  heat  is  required  to  produce  this  steam  from 
feed  water  at  a  temperature  of  90°  Fahrenheit  ? 

Solution.  The  total  heat  contents  \i  of  one  pound  of  steam  under  the  above  con- 
ditions is  AI  =  q\  +  TI  +  hi. 

From  Table  3,  qi  =  321.5,  TI  =  869.8,  and  hi  from  Problem  35  is  42  B. t.u.  Hence 
Xi  =  1233.3  B.  t.  u. 

For  feed  water  at  90°  Fahrenheit,  X2  =  g2  =  58.1  B.  t.  u. 

Hence  per  pound  Xi  —  X2  =  1175  B.  t.  u.  or  2350  B.t.  u.  for  2  pounds. 

37.  Entropy.     Find  the  entropy  for   the   superheated  steam  of 
Problem  36,  and  also  the  increase  in  entropy  above  that  of  the  feed 
water. 

Solution.     The  entropy  of  superheat  is 


2 
From  Table  3,  0,t  =  1.0741,  0Wi  =  0.5037. 

Hence  0i  =  1.6266. 

For  the  feed  water  02  =  0.1117. 

Hence  0i  -  02  =  1.5149  or  3.030  for  2  pounds. 

38.  Quality.     If  the  entropy  of  steam  at  135  pounds  absolute  is 
1.60,  find  its  quality  approximately. 

Solution.     Using  the  values  of  Problem  37,  if  0i  =  1.60,  then 

<i>Sl  1^1  0i  —  0tvi  —  0vi  ~~  u.O^ol. 
Assuming  the  specific  heat  as  0.48,  we  have  hi  =  0.48(0i  —  00). 

And  =  2C°-48)^-*o)  =0.0229. 

#0  +  01 

Hence  0.96  0i  -  0.96  00  =  0.023  00  +  0.023  0i. 

Substituting  6»0  =  350  +  460  =  810  as  in  Problem  37, 

0i  =  °-983(8l°)  =  850,  corresponding  to  40°  of  superheat. 
0.937 

39.  Quality  after  adiabatio  expansion.     If  the  steam  of  Problem 
36  expands  adiabatically  to  3  pounds  back  pressure  (absolute),  find 
the  quality  of  the  mixture. 

Solution.     From  Problem  37,  0i  =  1.627. 

From  Table  3,  for  a  pressure  of  3  pounds,    0W2  =  0.201, 

and  0flf  =  1.684. 

By  equation  64,  ft^j..,,  1.087 -0.201 

1  1.684 


102 


THERMODYNAMICS 


40.  Intrinsic  energy.     Find  the  intrinsic  energy  of  the  steam  of 
Problem  36. 

Solution.     By  equation  63,  the  intrinsic  energy  ei  is 

Aei  =  qi  +  TI  +  hi  —  Api(vi  —  <r)  =  Xi  —  Api(v\  —  <r). 

From  Problem  33,  the  specific  volume  is  3.75  cubic  feet,  and  from  Problem  36  the 
total  heat  is  \i  =  1233. 
Therefore 

Aei  =  1233  -  (185)(Q144)(3.75-(r)=1139  B.  t.  u.  per  pound, 

77o 
or 

61  =  (778)  (1139)  =  885,000  foot  pounds  per  pound  of  steam. 

For  2  pounds  the  energy  is  1,770,000  foot  pounds. 

41.  Work  of  adiabatic  expansion.      Find  the  work    done  in  the 
adiabatic  expansion  of  Problem  39. 

Solution.     For  state  1,          pl  =  135,     ti  =  425,  ei  =  885,000  ft.  Ib. 

For  state  2,  p%  =  3, '     x2  =  0.846,          q2  =  109.6,  P2  =  946.4, 

1 

and  hence 


€2  =  -  (02  +  £2/02)  =  (778)  (911)  =  707,000  ft.  Ib. 
A 

The  work  of  the  adiabatic  expansion  is 

w  =  fi  -  c2  =  885,000  —  707,000  =  178,000  foot  pounds  per  pound. 
For  2  pounds  of  steam  the  work  is  356,000  foot  pounds. 

42.    Grain  by  superheating.     The    steam   at   cut-off   in  a  certain 
engine  cylinder  has  a  quality  of  80  per  cent.     It  is  found  that  super- 


228° 


3ac 


0.336    0.475 


FIG.  51 


1.378 


1.603 


heating  the  steam  120  degrees  makes  the  quality  unity  at  cut-off. 
The  entering  steam  is  at  a  pressure  of  100  pounds  absolute.     The 


SUPERHEATED   STEAM 


103 


back  pressure  is  20  pounds  absolute.  The  specific  heat  may  be 
taken  at  0.52.  What  is  the  gain  in  efficiency  due  to  superheating? 
Plot  the  temperature  entropy  relations  for  both  cases.  Assume 
adiabatic  expansion. 

Solution.     The  0-<f>  plots  for  the  condition  of  80  per  cent  quality  and  100  per  cent 
quality  are  shown  to  scale  in  Figures  51  and  52  respectively.     The  useful  work  is  repre- 


448°  F 


328' 


228 


32' 


0.336     0.475 


1.602    1.677 


FIG.  52 


sented  in  heat  units  by  the  doubly  cross-hatched  area  in  each  case.     The  entire  cross- 
hatched  area  represents  the  heat  supplied. 

The  useful  work  qx  —  q2  is  found  numerically  by  assuming  the  water  line  oo1  a 
straight  line  and  writing 

qi  -  q2  =  [K0WJ  -  0«-,)  +  Si0*J(0i  ~  fts). 
The  useful  work  for  80  per  cent  quality  is  then 

qi  _  q2  =  [1(0.4748  -  0.3362)  +0.80(1.  1273)]  (327.9  -  227.9)  =  97.1  B.t.u. 
Similarly,  for  100  per  cent  quality, 

q1'-q2'  =  119.7  B.t.u. 

The  input  is  qi  =  qi  +  n—  g2  =  298.5  +  887.6  -  196.4  =989.7  B.t.u.  for  the  first 
case.  For  the  superheated  steam  <\if=qi  +  ri+hi  —  q^wherehi  is  approximately  (0.52) 
(120)  or  62.4  B.  t.u.  The  input  in  the  second  case  is  therefore  1186+62  =  1052  B.  t.u. 


The  efficiency  is  then 


=  9.81  per  cent  for  wet  steam  and 


=  9.60  percent 


989.7  1052 

for  dry  steam  at  cut-off. 

For  the  conditions  of  the  present  problem  the  gain  in  efficiency  due  to  superheating 
sufficiently  to  avoid  cylinder  condensation  during  admission  is  1.54  per  cent  in  9.81 
per  cent  or  an  increase  of  the  efficiency  of  15.7  per  cent. 


CHAPTER  Y 


FLOW   OF    STEAM   AND   GASES 

64.    Formula  of  de  Saint- Venant.     Consider  a  compressible  body 
B  (Figure  53)  acted  upon  by  forces  F1  and  F^.     Let  dz1  represent 

the  distance  of  motion  of  the  point  of 
application  of  Fv  arid  similarly  dz^  rep- 
resent the  motion  of  the  point  of  ap- 
plication of  FI  .  The  motion  dz^  is  of 
course  negative  if  there  is  expansion  as 
shown  in  the  figure.  Then  the  gain 
in.  energy  of  the  entire  mass  B  is 
equal  to  the  algebraic  sum  of  the  work  of  the  acting  forces,  or 


dz, 

dz', 

dZ2 

B 

B' 

> 

F; 

— 

If  we  now  consider  two 
adjacent  elements  B  and 
B1  of  an  aeriform  body, 
as  shown  in  Figure  54,  it 
is  evident  that  the  force 
FII  acting  upon  the  ele- 
ment 5,  is  equal  and 
opposite  to  the  force  F2f  acting  upon  Br.  Hence,  for  this  case,  the 
total  work  done  upon  the  entire  body,  B  plus  B ',  is  given  by  the 
expression 


FIG.  54 


This  is  the  gain  in  energy  of  the  entire  mass,  and  will  be  repre- 
sented in  subsequent  discussions  by  the  symbol  dO.  The  total 
change  for  an  extended  body  will  evidently  be  an  expression  of  the 
same  form  as  (i). 

Consider  a  steady  frictionless  flow  of  steam  in  a  non-conducting 
channel  (7,  shown  in  Figure  55.  The  transformation  of  the  steam 
will  then  be  isentropic.  Let  the  cross-section  at  A  be  denoted  by 
al  and  at  B  by  «2,  the  corresponding  pressures  on  these  areas  by  p^ 
and  p2.  Then,  if  F-^  and  F2  represent  the  forces  acting  upon  a1  and 
a2  respectively, 

Fl=plal    and    F2=p2a2 (ii) 

104 


FLOW   OF   FLUIDS 


105 


Since  the  flow  is  steady,  there  is  the  same  amount  of  steam  in 
pounds  passing  A  as  is  passing  B  in  the  same  time.  If  ^represents 
the  weight  of  steam  pass- 
ing A  or  B  in  one  second, 
and  if  the  velocities  at 
A  and  B  are  respectively  t^ 
and  t)2  in  feet  per  second, 
then  this  condition  of 
steady  flow  may  be  ex- 
pressed as 


and         Mv2  =  #2t>2, 


where    v1   and   v%    are    the 


FIG.  55 


specific     volumes     of     the 
steam  in  the  states  in  which  it  passes  A  and  B  respectively.     Equat- 
ing M  from  the  two  equations  of  (iii)  gives 


(65) 


This  equation  is  known  as  the  "  equation  of  continuity,"  and  holds 
for  any  fluid  in  a  condition  of  steady  flow. 

Substitution  of  this  equation  of  continuity  in  the  equation  (i)  for 
the  increase  of  energy  gives  an  equation  expressing  this  increase  dO 
in  terms  of  the  pressure  and  specific  volumes  at  A  and  B.  Thus,  if 
we  consider  the  flow  for  a  small  fraction  of  time  c?t,  then 


dz1  =  fcjcft   and 


(iv) 


In  the  time  dt  the  mass  of  steam  passing  A  or  B  may  be  expressed 

o  c 

dm  =  Mdt  .........     (v) 

Hence,  for  this  mass,  dm,  the  equation  of  continuity  (48)  becomes 


, 

dm  = 


,   ., 
(vi) 


Substitution  from  equations  (ii),  (iv),  (v),  and  (vi)  in  equation  (i) 

gives 

dO  =  M(plvl-p^)dt.      .....     (vii) 

The  total  increase  of  energy  expressed  by  equation  (vii)  is  partly 
external  kinetic  energy  of  the  entire  mass  of  steam,  Mdi,  and  partly 


106 


THERMODYNAMICS 


intrinsic  or  molecular  energy.     Representing  these  components  by 
dK  and  dE  respectively,  we  have 


dO=dK+dE. 


The  external  kinetic  energy  expressed  in  terms  of  the  gain  in  veloc- 
ity of  the  entire  mass  of  the  steam  is 


dK= 


Mdt 


The  increase  of  intrinsic  energy  dE  is  equal  to  the  external  work 
done  in  compressing  the  steam.  If  there  is  expansion,  as  in  the  case 
considered,  the  increase  in  intrinsic  energy  is  negative,  and  is  numeri- 
cally equal  to  the  work  done  by  the  gas  in  expanding.  Thus, 


(x) 


Or  dividing  by  Mdt  and  transposing,  we  have 

2 


Substitution  from  equations  (viii),  (ix),  and  (x)  in  equation  (vii) 
gives 

(xi) 


(xii) 


From  Figure  56,  which  represents 
the  pressure-  volume  plot  of  the 
frictionless  adiabatic  expansion 
under  consideration,  it  is  evident 
that  p1vl  is  equal  to  the  area  Adog, 

CV2 

p2v%  to  the  area  Scof,  and     I    pdv 

to  the  area  ABcd.     Hence  equation 
(xii)  becomes 


'=  area  ABfg=^vdp,  (66) 


This  equation  may  be  expressed 
in  terms  of  the  pressures,  and  spe- 
cific volumes  at  A  and  B  if  the  frictionless  adiabatic  expansion 
may  be  taken  as  represented  by  an  equation  of  the  form 


FLOW   OF   FLUIDS  107 

Then 

p  ,  .   , 

(XIV) 


n-  lv 
Therefore 

E22  -  ttta  =     rc     f 
2#         n-l      1  l 

In  this  equation,  which  is  due  to  de  Saint- Venant,1  p  and  v  are 
specific  pressures  and  specific  volumes  in  pounds  per  square  foot 
and  cubic  feet  per  pound,  respectively.  The  kinetic  energy  term 
represents  the  increase  in  external  kinetic  energy  per  pound  of 
steam.  The  exponent  n  is  given  by  Zeuner  as  1.035  4-  0.1  x  for  wet 
steam  of  quality  x\  that  is,  1.135  for  steam  initially  dry,  and  n  is 
1.3  for  superheated  steam. 

•  65.  Alternative  Expression  for  the  Formula  of  de  Saint-Venant. 
If  in  equation  (viii),  of  the  preceding  section  we  substitute  for 
dK  from  equation  (ix)  of  the  same  section,  but  for  dE  write 
(e2  —  €j)  Mdi,  where  e:  and  e2  are  the  values  of  the  intrinsic  energy 
per  pound  of  steam  at  A  and  at  B  respectively,  we  have 

2)dt  =  Mdi 

or  dividing  by  Mdt  and  transposing,  we  obtain  for  a  pound  of  a 
mixture  undergoing  a  frictionless  adiabatic  expansion  the  relation 

i  .     P^       .  .     P2  /'£\Q\ 

11       l      2g        2  2       2      Zg 

Equations  (66)  and  (67)  of  Section  64  and  equation  (68)  given 
above  as  well  as  the  equation  of  continuity  (65)  hold  equally  well 
for  all  fluids.  Thus  if  the  fluid  is  a  gas,  the  value  of  the  exponent 

n  is  of  course  K  —  ^  =  1.4  for  diatomic  mixtures, 


(69) 


If   the  fluid  is  incompressible  as  water,  then  vl  =  v2  and  for  an 
adiabatic  change  ej  =  e2, 

t)  2—  t)  2 
whence  -2-j  —  L  =  (p^  —  p%)  cr  ........      (70) 

•^  y 

1  A  more  rigorous  demonstration  from  which  the  above  is  derived  is  to  be  found  in 
Lowenstein's  translation  of  Stodola,  "Steam  Turbines,"  2d  ed.,  pp.  4-8. 


108  THERMODYNAMICS 

For  steam  three  cases  may  be  distinguished  ;  namely,  (1)  expan- 
sion through  the  channel  from  a  condition  of  saturated  steam  of 
quality  xl  to  a  condition  of  quality  x2,  (2)  expansion  from  a  super- 
heated condition  to  a  saturated  condition  of  quality  o?2,  and  (3)  ex- 
pansion entirely  within  the  superheated  region.  For  all  these  cases 
the  expression  for  the  gain  in  kinetic  energy  per  pound  of  the  steam 
mixture  reduces  to  a  simple  form  which  will  be  derived  in  the  fol- 
lowing section. 

66.  Zeuner's  Formula  for  Kinetic  Energy.  In  equation  (68) 
transposition  gives 


Considering  an  expansion  which  takes  place  entirely  within  the 
region  of  wet  steam,  we  have 


vl  =  x^  -4-  cr,  and  v2  =  x2u2  +  cr. 


Therefore,  neglecting  cr,  which  in  general  will  make  a  difference  of 
less  than  half  of  one  per  cent  and  substituting  from  equation  (i)  in 
equation  (68),  we  may  write 


But 


\)  -  02  +  Saffa11*)  =  ^       t?1 


and  ^  = 


Substituting  from  equation  (iii)  in  equation  (ii)  and  remembering 
that 

\l  =  q1  +  xfa  +  x^Ap^  and  X2  =  q2  +  z2/>2  +  x^Ap^u^   .      (iv) 
gives 


Similarly,  consider  the  steam  superheated  in  state  1  and  wet  in 
state  2.     For  superheated  steam,  equation  (63)  of  Section  60  gives 


Or  neglecting  cr  as  before, 

fe  +    +       -tfXH-     •    •       •  (vi> 


FLOW   OF   FLUIDS  109 


Substituting  for  (CI+PI»^)  from  equation  (vi)  and  for 
from  equations  (iii)  and  (iv)  in  equation  (ii)  gives 


For  the  case  of  an   expansion   entirely  within  the  superh'eated 
region  similar  reduction  leads  to  the  same  equation  as  (71). 

67.  Kinetic  Energy  of   an  Irreversible  Adiabatic  Expansion.     If, 
however,  the  decrease  in  total  heat  contents  is  not  entirely  converted 
into  external  energy,  but  is  partly  expended  in  external  work  of 
amount  w,  while  some  heat  of  amount  q  is  transferred  from  the 
substance  to  external  bodies,  we  have  to  consider  the  general  case  of 
an  expansion  with  heat  changes  and  external  work.     Then  equa- 
tion (71)  becomes 

\  -  X2  =  A  *a2J~  *\  +  q  +  Aw.      ....     (72) 

t/ 

If  there  is  work  done  against  friction,  there  must  be  added  to  the 
right-hand  side  of  equation  (71)  a  term  y  representing  the  work  per 
pound  done  against  friction  and  expressed  in  heat  units.  If  q  is 
zero,  then  the  expansion  is  adiabatic  ;  if  w  is  zero,  then  there  is  a  free 
expansion  ;  but  if  y  is  not  zero,  then  there  is  an  adiabatic  expansion 
similar  to  that  described  in  the  case  of  the  "porous  plug  experiment  " 
of  Section  35,  which  is  irreversible  and  during  which  there  is  an 
increase  in  entropy.  This  case  of  the  irreversible  expansion  of  steam 
in  a  non-conducting  channel  will  be  discussed  in  the  latter  part  of 
the  following  section. 

68.  Loss  of  Kinetic  Energy  Due  to  Friction.     If  there  is  friction 
during  the  expansion  from  a  pressure  of  pl  to  a  pressure  of  p^  the  gain 
in  kinetic  energy  will  be  less  than  it  would  have  been  for  a  similar 
frictionless  expansion.     Denote  by  t)2'  the  final  velocity  for  a  How 
with  friction   for   the   same   range   of   pressures  as  gives   D2  for  a 
frictionless  expansion.     Then  denoting  by  I  the  loss  of  kinetic  energy 
per  pound  of  the  mixture  expressed  in  thermal  units,  we  have 

A*t-*i*.  •     •     •     (73) 
' 


Denoting  by  X2'  the  total  heat  contents,  after  flow  with  friction,  it  is 
clear  that 

Z  =  (^i-*<2)-04-V)  =  V->2  .....  (74) 
where  \  and  X2  have  their  former  significance  and  refer  to  a  friction- 
less  flow. 


HO  THERMODYNAMICS 

As  the  expansion  takes  place,  a  portion  of  the  kinetic  energy  of 
the  mixture  is  converted  into  heat  by  friction  and  immediately  after- 
wards taken  up  by  the  mixture  at  a  lower  temperature  and  availa- 
bility and  again  converted  into  kinetic  energy.  Since  the  heat 
equivalent  of  the  friction  work  is  thus  continually  being  returned  to 
the  mixture,  the  actual  loss  in  external  kinetic  energy  due  to  the 
friction  is  less  than  the  total  work  done  against  frictional  forces. 
The  heat  corresponding  to  this  friction  work  has  been  represented 
in  the  previous  section  by  y. 

This  heat  y  when  returned  to  the  steam  or  fluid  in  successive 
infinitesimal  amounts  dy  produces  changes  in  the  intrinsic  energy  of 
the  mixture  of  amount  to  be  represented  by  de  and  also  produces  an 
amount  of  external  work  of  value  pdv.  That  is, 

dy  =  Ade  -f  Apdv.    . (i) 

But                      Ad(pv)  =  Apdv  +  Avdp (ii) 

Hence                          dy  =  Ade  +  Ad{pv)  —  Avdp.       .     .     .  (iii) 
In  equation  (iii)  there  may  be  substituted  as  in  Section  66 

Ade  +  Ad(pv)  =  A(e1+p1v1)-A(e2+p2v2)=d\.   .  (iv) 

Hence                         dy  =  d\-Avdp (75) 

By  integration  from  state  1  to  state  2 


•  •  <76> 

where  equation  (76)  is  a  general  relation  for  the  total  amount  of 
friction  work. 

If  now  the  expansion  is  frictionless  and  takes  place  from  state  1  to 
state  2,  we  have  y  =  0  and 

X2  —  Xx  =  A  I    *vdp.      .  (v) 

•ffi 

For  the  expansion  with  friction  between  the  states  1  and  2'  at 
which  the  pressure  is  the  same  as  for  state  2  but  at  which  the  volume 

vr  is  different,  />2' 

y  =  \2f  —  Xj  —  A  I     v'dp.      .  .     .      (vi) 

Subtracting  equations  (v)  and  (vi)  gives 

i'l  A(  Cp*  ?i       Cp*  *  \ 

y  =  \2'  _  X2  —  A  f   I     v'dp  —  I     vdp  J, 

or, 

rpa 

•     •     (vii) 


FLOW   OF   FLUIDS 


111 


In  Figure  57  is  shown  the  p-v  plot  for  these  two  expansions. 
From  this  it  is  evident  that 


JO- 


area  122', 


FIG.  57 


or  reversing  the  order  of  the  integration  and  thus  changing  the 

algebraic  sign, 


PL 

Therefore 


_  CP\VI'  -  v)dp  =  +  CP\vr  -  v)dp  =  area  12'2. 

*Sn.  Pz 

y  =  \2f  -  \j  +  A  x  area  12'  2 (77) 


The  total  work  done  against  friction  is  therefore  larger  than  the 
loss  of  kinetic  energy  due  to  friction  by  the  amount  of  work  cor- 
responding to  the  area  on  the  p-v  plot  included  by  the  line  repre- 
senting the  actual  expansion,  the  line  representing  an  adiabatic 
expansion,  and  the  line  of  constant  final  pressure. 

The  temperature  entropy  plot  for  these  expansions  is  shown  in 
Figure  58.  For  a  frictionless  adiabatic  expansion  the  area  ab!2 
represents  in  heat  units  the  increase  in  kinetic  energy  per  pound. 
If  all  the  kinetic  energy  of  the  expansion  were  converted  into  fric- 
tion and  this  friction  into  heat,  the  line  13  representing  the  curve  of 
constant  total  heat  contents  through  1  would  be  the  path  followed 
by  the  state  point  of  the  steam  mixture.  The  area  23fd  would 
then  be  equal  to  the  area  ab!2.  The  path  followed  by  the  state 
point  during  the  actual  irreversible  expansion  through  a  channel 
like  a  turbine  nozzle  is  some  line  between  the  constant  heat  line  13 
and  the  reversible  adiabatic  12,  such  as  12' .  For  the  path  12r  the 


112 


THERMODYNAMICS 


e 


friction  work  is  12'ed,  of  which  22' ed  represents  the  loss  in  kinetic 
energy  I.  The  net  increase  of  kinetic  energy  during  an  irreversible 

adiabatic  expansion  is  then 
expressed  in  heat  units  by  the 
area-  12ab  diminished  by  the 
area  22' ed. 

The  loss  I  is  comparatively 
small.  For  nozzles  such  as 
are  used  in  turbines  which  are 
not  more  than  2  inches  long 
I  is  about  5  to  8  per  cent  of 
the  energy  which  would  be 
developed  in  a  friction  less 
flow.  For  larger  nozzles  of 
length  4  to  6  inches  and  with 
diameters  at  the  narrowest 
point  of  from  one  quarter  to 
three  eighths  of  an  inch  I  is 
from  10  to  15  per  cent.1 

69.  Flow  through  an  Orifice. 
If  a  fluid  flows  from  a  reser- 
voir of  such  size  that  t)x2  may 
be  neglected  in  comparison 
with  t>22  (as  is  the  case  in  the 
flow  of  steam  from  a  boiler 
into  a  pipe),  then  equations 
(71)  and  (65)  may  be  used 
to  find  the  weight  of  fluid 
passing  through  the  orifice  if 
the  second  state  be  taken  at 
the  narrowest  point  or  throat 

of  the  short  tube  or  orifice.  The  reason  for  specifying  that  the 
second  state  shall  be  that  at  the  throat  is  evident  from  the  following 
experimental  fact  and  also  may  be  seen  from  the  curves  of  Prob- 
lem 53. 

Experiments  of  Napier  show  that  the  weight  of  steam  discharged 
through  an  orifice  increases  for  a  constant  pressure  p^  with  a  reduc- 
tion of  the  back  pressure  p2  until  p2  is  Q.58pv  after  which  further 
reduction  does  not  increase  the  weight  discharged.  In  other  words, 


FIG.  58 


1  The  discussion  follows  Stodola,  "  Steam  Turbines,"  pp.  48-63. 


FLOW   OF   FLUIDS  113 

the  pressure  in  the  orifice  for  steam  is  always  such  as  to  give  a 
maximum  value  for  the  weight  of  steam  discharged  per  second. 
For  steam,  therefore,  the  pressure  at  the  orifice  may  be  taken  as 
0.58^!  provided  that  p1  >  1.73  p2. 

Rankine  therefore  announced  from  his  study  of  the  experimental 
results  obtained  by  Napier  the  following  empirical  formulas  for 
calculating  the  flow  of  steam  through  an  orifice.  The  symbols  have 
the  significance  of  the  preceding  discussion. 

Where  p^  ^  -J  jt?2, 


and 

where  p1  < 

"^ (79) 


It  is  to  be  noticed  that  these  formulas  of  Napier  and  the  assumption 
of  the  pressure  in  the  throat  as  0.58^  are  only  approximations. 
Since  no  flow  is  rigorously  frictionless  or  adiabatic  and  the  variations 
from  these  ideal  conditions  depend  upon  the  orifice  or  tube  employed 
when  greater  accuracy  than  about  2  per  cent  is  required,  the  given 
nozzle  should  be  calibrated  by  weighing  the  steam  passing  through 
it  for  various  values  of  p^  and  p2. 

70.  Fliegner's  Formulas  for  the  Flow  of  Air.  For  the  flow  of  air 
through  a  rounded  orifice,  where  the  pressures  are  p1  and  p2  on  the 
two  sides  of  the  orifice  and  the  area  is  a\  the  experiments  of  Fliegner 
show  that  the  weight  M  of  air  in  pounds  per  second  is  given  by  the 
following  formulas  : 

where  p1  >  2  p2 

M  =  0.530  a-^=;        •     (80) 

V0j 

where  p1  <  2  p^ 

pi^E2 (81) 

In  these  formulas  0l  is  the  absolute  temperature  Fahrenheit  of  the 
air  in  the  reservoir.  The  pressure  and  the  area  are  to  be  measured 
in  any  similar  units,  as  either  in  pounds  per  square  foot  and  in  square 
feet,  or  the  pressures  in  pounds  per  square  inch  and  the  area  in 
square  inches. 


114  THERMODYNAMICS 

71.    Resume  of  Equations  for  Flow  of  Fluids. 
Equation  of  continuity. 

//n        /»  n 

(65) 


Frictionless  adiabatic  flow  of  steam. 
t)  2—  t)  2 

Greneral  equation. 

Friction  work. 

y  =  I  4-  A  (area  between  isentropic  and  actual  expansion  paths 

on  a  p-v  plot) (77) 

Loss  of  external  K.  E.  due  to  friction. 

l  =  \J-\. (74) 

Napier's  formulas  for  steam  flow. 

#"=«&,  where  ft  >|ft (78) 


M=  a-i,  where  ft  <   ft.        .     .     (79) 


Fliegner^s  formula  for  air  flow. 

M=  0.530  a-^L,  where  p1  >  2p2  .....     (80) 


M  =  1.060  a\J*i~,  where  p1  <  2p^.     .     (81) 
#1 

PROBLEMS   AND    SOLUTIONS 

43.  Kinetic  energy  of  steam.  Given  one  pound  of  steam  initially 
dry,  issuing  from  a  large  reservoir  under  a  pressure  of  100  pounds 
absolute.  Assume  adiabatic  expansion  to  a  back  pressure  of  14.7 
pounds  absolute.  Consider  the  velocity  in  the  reservoir  negligible. 
Find  the  final  velocity  (a)  using  equation  (67),  (5)  using  equation 
(71);  (tf)  find  the  error  due  to  neglecting  the  term  0(p1  — 


FLOW   OF   FLUIDS  115 

Solution.  From  Table  3,  for  100  pounds,  q{  =  298.5,  r\  =  887.6,  <f)w  =0.4748, 

0Wi  =  1.1273,  81  =  4.432;  for  _p2  =  14.7  pounds,  q2  -  180.3,  r-2  =  969.7,  </>  J  =  0.3125, 
<t>vl=  1.4441,  s2  =  26.78. 

Hence  the  quality  after  adiabatic  expansion  without  friction  is 

_  0U,,  +  4s  -  Is  _  0.4748  +  1.1273  -  0.3125 
~^7~  1.4441 

(a)  t?i  =  si  =  4.432  ;  v2  =  X2s2,  approximately,  or  (0.893)  (26.78)=  23.92. 
By  equation  (67), 

^-2  =  1^  [(100)  (144)  (4.432)  -  (14.7)  (144)  (23.92)  ]=  111,100  foot  pounds. 
2  ^      0.135 

Therefore  ba  =  2670  feet  per  second. 

(6)  Xi  =  298.5  +  887  .6  =1186.1,   X2  =  180.3  +  jr2(969.7)=  1046.4. 
Hence,-  in  equation  (71),     «7(\i  —  X2)  =  108,800  foot  pounds, 

the  term  o-(pi  —  #2)  is  of  value  (0.016)  (144)  (100  —  14.7)=  196  foot  pounds,  and  is 
negligible. 

Therefore,  neglecting  this  term,  we  have 

^  =  108,800, 

or  D2  =  2650  feet  per  second. 

(c)  The  term  neglected  above  evidently  introduces  an  error  of  0.2  per  cent  in  the 
value  of  the  kinetic  energy,  or  0.1  per  cent  in  the  value  of  the  final  velocity. 

44.  Kinetic  energy  of  superheated  steam.  Given  one  pound  of 
steam  at  100  pounds  absolute  and  120°  Fahrenheit  of  superheat, 
find  the  final  velocity  acquired  in  expanding  along  a  reversible  adia- 
batic to  a  back  pressure  of  14.7  pounds  absolute,  (a)  by  using  equa- 
tion (67),  (6)  by  using  equation  (71),  as  in  Problem  43. 

Solution.  From  Table  3,  the  values  of  the  constants  for  these  pressures  may  be 
obtained,  as  given  in  the  preceding  problem.  For  120°  of  superheat  cp  may  be  taken 
as  0.53.  Then 

hi  =  cp(0i  —  00)  =  0.53(120)  =  63.6  B.  t.  u., 


The  specific  volume  is  found  by  equation  (58)  to  be  5.27  cubic  feet. 
The  quality  after  expansion  is,  then, 

0«!  +  <t>*i  +  0i  -  0«3     0.4748  +  1.1273  +  0.0750-0.3125  _  Q  Q45 
0C2  1.4441 

(a)  Since  the  expansion  extends  from  the  superheated  into  the  saturated  region,  it 
is  necessary  to  divide  the  expression  of  equation  (67)  for  the  increase  of  kinetic  energy 
into  two  parts,  one  representing  the  increase  while  the  steam  is  superheated  and 
calculated  by  using  n  =  1.3,  and  one  representing  the  increase  while  the  steam  is  satu- 
rated and  calculated,  as  in  Problem  43,  by  using  n=  1.135.  This  demands  a  knowl- 
edge of  the  pressure  and  volume  corresponding  to  the  intersection  of  the  isentropic  line 


116 


THERMODYNAMICS 


with  the  curve  of  constant  steam  weight.  This  is  most  easily  found  from  the  0-0  chart 
of  page  86  to  be  40  pounds  for  the  pressure  and  10.5  cubic  feet  for  the  specific  volume. 
It  may  be  found  approximately  by  using  the  values  for  the  entropy  at  30  and  50 
pounds,  and  assuming  that  the  rate  of  change  of  the  entropy  is  proportional  to  the 
rate  of  change  of  the  pressure.  Such  an  interpolation  gives  the  pressure  as  39  pounds. 
The  more  accurate  value  of  40  pounds  will  be  used  in  this  solution.  Substitution  of 
these  values  in  equation  (67)  then  gives 

|£  =  i|  [  (100)  (144)  (5.27)  -  (40)  (144)  (10.5)  ] 

+  j^||  [  (40)  (  144)  (10.5)  -  (14.7)  (144)  (25.3)] 

=  67,000  +  58,100  =  108,000  foot  pounds. 
Hence  t>«  =  2830  feet  per  second. 

(6)  Solution  by  equation  (71)  is  similar  to  that  of  Problem  43  except  that  \i  is 
greater  by  the  superheat  hi. 

\i  =  1186.1  +  63.6  =  .1249.7,  X2  =  180.3  +  (0.945)  (969.7)  =  1096.5. 


-  X2)  =  778(153.2)  =  119,200  foot  pounds. 


Hence  t)2  =  2770  feet  per  second. 


Discussion. 


320°  F 


300° 


280° 


260° 


240 


220 


200° 


The  deviation  between  solutions  (a)  and  (&)  is  due  to  the  inaccuracy 

with  which  n  is  accompanied. 
The  method  of  (&)  is  the  more 
accurate  and  in  general  shorter  of 
solution. 


1.60 


1.62  1.64 

FIG.  59 


1.66 


1.68 


45-  Loss  due  to  friction. 
In  Problem  43  the  ve- 
locity for  a  frictionless 
adiabatic  expansion  has 
already  been  found.  Tak- 
ing the  same  values  for  the 
pressures  but  assuming  a 
loss  of  16  per  cent  of  the 
energy  available  for  in- 
creasing .  the  kinetic  en- 
ergy, find  the  final  velocity. 
Sketch  the  0-$  plot  and 
find  the  total  work  done 
against  friction.  Assume 
the  plot  of  the  actual  ex- 
pansion line  on  the  0-<f> 
diagram  to  be  a  straight 
line. 

Solution.  The  increase  in 
kinetic  energy  of  the  steam  for 


FLOW    OF   FLUIDS  117 

a  reversible  adiabatic  has  been  determined  as  108,800  footpounds,  and  the  final  quality 
as  0.893.  The  loss,  I,  in  kinetic  energy  is  then  T^%(108,800)  =  17,400  foot  pounds 
=  22B.t.  u.  The  absolute  temperature  corresponding  to  the  back  pressure  is 
212  +  460=672  degrees.  The  increase  in  entropy  due  to  the  friction  is  then  $fy  = 
0.0328  and  is  represented  in  Figure  59  by  the  line  22'. 

The  area  12'2  may  now  be  found  approximately  by  multiplying  half  this  increase  in 
entropy  by  the  difference  in  temperature  between  points  1  and  2  ;  that  is, 

area  12'2  =  i(0.0328)(328  -  212)=  1.9,  or  approximately  2  B.  t.  u. 

The  work  done  against  friction,  y,  is  by  equation  (77)  equal  to  the  sum  of  this  area 
and  I  Hence  y  =  22  +  2  =  24  B.  t.  u.,  or  18,700  foot  pounds. 

The  final  velocity  is  of  course  one  half  of  16  per  cent  smaller  than  that  for  a  fric- 
tionless  flow,  since  the  energy  available  is  16  per  cent  smaller.  Hence  t)2  =  24400  feet 
per  second. 

46.  Flow  of  air.  Find  the  weight  of  air  passing  per  second 
through  a  rounded  orifice  of  area  0.1  square  inch,  if  the  pressures  on 
the  two  sides  of  the  orifice  are  45  and  14.7  pounds  absolute,  (a)  by 
equation  (67),  (6)  by  equation  (80).  The  temperature  of  the  air 
within  the  reservoir  is  80°  Fahrenheit. 

Solution,  (a)  The  pressure  pt  in  the  throat,  where  the  area  is  given  above,  is  to  be 
taken  according  to  Fliegner's  experiments  as  0.577  of  p±,  or  for  this  case  pt  =  (0.577)  (45). 
The  specific  volume  of  the  air  in  the  reservoir  is,  by  equation  (13), 

=  80  +  460  14.7  12  39  = 
460        45 


The  specific  volume  after  expansion  is,  by  eq.  (32), 

(4.75)  =  7.02  cubic  feet. 


=  [; 


i 
45       I1-* 


.  (0.577)  45  J 
Hence,  by  eq.  (65), 


3/=—^^ ^1^2  ^[(45) (144) (4.75) -(0.577) (144) (7.02)]  =  0.100    pounds   per 

second. 

(6)  By  equation  (80),  since  61  =  80  +  460  =  540, 

M  =  (0.530)  (Q.  1)    45    =  0.102  pound  per  second. 
V540 

47.  Napier's  formula.  Given  steam  of  95  per  cent  quality  expand- 
ing from  120  pounds  absolute  to  atmospheric  pressure  of  14.7  pounds 
through  an  orifice  of  0.1  square  inch  area.  Assume  the  pressure  in 
the  throat  to  be  0.58^  or  70  pounds  absolute.  Find  the  weight  of 
steam  per  second,  (a)  by  equation  (71),  (5)  by  equation  (78). 

Solution,  (a)  From  Table  3,  for  120  pounds,  gi  =  312.3,  n  =  876.9,  0Wj  =0.4922, 
0^  =  1.0951,  For  pt  =  70  pounds,  qt  =  272.9,  rt  =  906.6,  0«t  =  0.4418,  0r<  =  1.1892, 
&  =6.199. 


118 


THERMODYNAMICS 


Hence 


=  0.4922 +  (0.95)^0951) -0.4418  = 


And 


\i  -  X,  =  312.3  +  (0.95)  (876.9)  -  272.9  -  (0.917)  (906.6)  =  41.1. 
Then 


» 


(0.1)  (120) 
70 


pound         second. 


48.  Napier's  formula.  Compare  the  weight  of  steam  of  quality 
95  per  cent  passing  through  an  orifice  0.1  square  inch  in  area  from 
an  initial  pressure  of  20  pounds  absolute  to  atmospheric  pressure  of 
14.7  pounds,  as  calculated  (a)  by  equation  (71),  (6)  by  equation 
(79). 

Solution,     (a)  From  Table  3,  for  pi=  20  pounds,  qi=  196.4,  7*1=  959.4, 
0Vi  =  1.3957.     For  p2  =  14.7  pounds,  q«  =  180.3,  r2  =  969.7,  <j>w2=  0.3125, 
s2  =  26.78. 
Hence 

X2  =  0.3362  +  (0.95)  (1.3957)  -0.3125  =  ^6  or  0.934. 

1.4441 

t?2  =  (0.934)  (26.78)  =  25.0. 

\!  _  \2  =  196.4  +  (0.95)  (959.4)  -  180.3  -  (0.934)  (969.  7)  =  21.9  B.t.u. 
Therefore 


i=  0.3362, 
v^=  1.4441, 


(6)      M  = 


second' 


49.  Injector  (non-lifting}.  An  injector,  as  the  name  implies,  is  a 
device  for  injecting  water  into  a  boiler  against  the  boiler  steam 
pressure.  This  is  accomplished  by  a  mechanism  the  principles  of 
which  are  evident  from  Figure  60.  To  start  the  injector,  steam  is 

Steam 


FIG.  60 


allowed  to  flow  from  the  steam  delivery  pipe  through  the  injector 
and  out  of  the  overflow  F,  thus  blowing  out  any  condensed  water 


FLOW   OF   FLUIDS 


119 


and  insuring  practically  dry  steam.  The  steam  valve  S  is  then 
turned  off  and  the  water  valve  turned  on.  When  the  water  appears 
at  the  overflow  the  steam  -valve  and  the  valve  in  the  delivery  tube 
to  the  boiler  are  both  turned  on.  The  entering  steam  is  condensed 
by  contact  with  the  colder  feed  water  and  converts  some  of  its  heat 
energy  into  m'echanical  kinetic  energy,  forcing  feed  water  and  the 
condensed  steam  into  the  boiler.  The  partial  vacuum  formed  by 
the  steam  and  water  rushing  by  the  overflow  valve  draws  it  shut. 

Calculate  the  number  of  pounds  of  water  forced  into  a  boiler 
against  a  pressure  of  120  pounds  absolute  by  one  pound  of  steam 
from  the  same  boiler.  Neglect  the  velocity  of  the  entering  water. 
Assume  the  steam  98  per  cent  dry,  the  temperature  of  the  water 
supplied  to  the  injector  as  70°  Fahrenheit,  and  the  temperature  of 
the  water  leaving  the  injector  as  170°  Fahrenheit.  Assume  that  the 
kinetic  energy  of  the  entire  mass  of  water  leaving  the  injector  is  one 
per  cent  of  the  total  heat  contents  of  one  pound  of  steam. 

Solution.  Let  there  be  ra  pounds  of  feed  water  entering  the  injector  for  every 
pound  of  steam.  The  total  energy  of  the  entering  water  plus  that  of  the  entering 
steam  must  equal  the  total  energy  of  the  two  combined  as  they  leave  the  injector. 
That  is, 

W\Water  +  K.  E.water  +  Xsteam  +  K.  E.steam  =  (1  -f-  m)X(water+steam)  +  K.  E.(water+  steam), 

where 

Xwater  =  39  B.  t.  u.  ;  K.  E.  water  is  negligible  ;  Xsteam  =  q  +  xr  =  1172  B.  t.  u.  per  pound. 
K.  E.(water+steam)  =  12  B.  t.  u. ;  and  X(water+steam)  =  138  B.  t.  u.  per  pound. 

Hence  39  m  +  1172  -  (1  +  w)138  +  12   or  m  =  10  pounds  approximately. 
For  practical  purpose  it  is  sufficient  to  neglect  all  kinetic 
energy  terms  and  write 

m  =  (rr)*+q.-qw+8  =  1172  -  138 
qw+s-qw  '    138-39 

=  10  pounds  approximately. 

50.  Peabody  throttling  calorimeter.  The 
Peabody  throttling  calorimeter  employs  the 
phenomena  of  an  irreversible  adiabatic  expan- 
sion in  the  determination  of  steam  quality. 
It  consists  of  a  reservoir  A  shown  in  Figure 
61  to  which  the  steam  of  which  the  quality 
is  desired  is  admitted  by  a  small  pipe  b.  A 
pressure  gauge  at  /  and  a  thermometer  at  e 
record  the  temperature  and  pressure  of  the 
steam  after  it  has  expanded  to  fill  the  reser- 
voir. The  valve  d  admits  of  the  free  passage 
of  steam  from  the  calorimeter.  The  entire  FIG.  61 


120 


THERMODYNAMICS 


reservoir  is  surrounded  by  a  heat  insulator.  For  a  test,  the  valve 
b  which  connects  the  reservoir  with  the  main  steam  pipe,  in  which 
it  is  desired  to  know  the  quality,  is  opened  slightly,  and  thus  admits 
steam  throttled  down  to  a  pressure  of  5  to  15  pounds  above  that  of 
the  atmosphere  to  which  d  is  connected. 

If  the  boiler  pressure  is  p\  =  80.3  pounds  gauge,  the  pressure  p2 
within  the  calorimeter  is  5.3  pounds  gauge,  the  atmospheric  pressure 
is  14.7  pounds,  and  the  temperature  within  the  calorimeter  is  278° 
Fahrenheit,  find  the  quality  of  the  steam.  Sketch  the  temperature 
entropy  diagram  for  this  process. 

Solution.  There  is  assumed  to  be  an  irreversible  adiabatic  expansion,  in  which  as 
fast  as  the  heat  of  the  entering  steam  is  converted  into  kinetic  energy  of  motion,  this 
energy  is  transformed  into  work  of  friction  and  eddy  currents,  and  is  returned  at  a 
slightly  lower  temperature  to  the  steam.  The  change  thus  takes  place  at  constant  heat 
contents  Hence  \i  =  X2.  If  the  steam  is  nearly  saturated,  the  constant  heat  line  will 
take  it  into  the  superheated  region  for  lower  temperatures  and  pressures.  Hence  we 

AI  =  gi  +  xiri  =  \2=q2+r2  +  cp(t  -  *<>), 

where  to  is  the  saturation  temperature  corresponding  to  the  pressure  p<>,  and  t  is  tlie 
temperature  as  recorded  by  the  thermometer  in  the  calorimeter. 

Hence,  using  0.48  for  cp  and  substituting  the  proper  values  for  q  and  r  from  Table  3, 
we  have 

x  =  ga  +  r2+cp(t  -  fr)  -  gi  =  196.4  +  959.4+0.48(278  -  228)  -  204.6  _  Q  9Q1 
ri  890.5 

The  temperature-entropy  plot  is  shown  in  Figure  62,  where  are  plotted  to  scale  the 
corresponding  values  of  the  entropy  as  found  from  Table  3  for  the  constant  steam 
weight  line,  and  by  calculation  for  the  curve  of  constant  heat  contents  equal  to  AI. 


F 
340° 

320* 
300' 
280° 
260° 
240 
220° 

9 

1 

\ 

NX 

\ 

5 

~r« 

'JSp 

fe* 

\ 

\ 

, 

\ 

\ 

7 

i 

—  - 

— 

•  — 

-  — 



— 

g 

\ 

L64  1.68  L72 

FIG.  62 


L76 


1.64  1.68 

FIG.  63 


To  illustrate  more  fully  the  difference  in  appearance  between  the  6-$  and  the 
p-(f>  diagrams  discussed  on  page  96,  there  is  shown  in  Figure  63  to  the  same  scale  as 
used  above  the  p-(f>  plot  for  this  process. 


FLOW   OF   FLUIDS  121 

51.  Limits  of  throttling  calorimeter.     If  the  valve  d  of  the  calo- 
rimeter described  in  Problem  50  is  connected  not  to  the  atmosphere 
but  to  the  exhaust  pipe  of  the  engine  so  that  the  back  pressure  is  3 
pounds  absolute,  find  the  largest  amount  of  priming  (lowest  value  of 
x)  that  can  be  measured  with  the  calorimeter  for  a  boiler  pressure  of 
85  pounds  absolute.     Neglect  probable  and  observational  errors  in 
the  thermometer. 

Solution.  At  the  back  pressure  of  3  pounds  the  steam  must  in  the  limiting  case  be 
just  at  the  point  of  being  superheated.  Hence  putting  cp(t  —  tx~)  equal  to  zero  in  the 
equation  for  xi  derived  in  Problem  50  and  substituting  for  q  and  r  from  Table  3, 
we  have 

Xl  =  109.6  +  1012.2-286.5  =  0.9S15. 

For  this  pressure  the  calorimeter  can  then  be  used  to  determine  the  steam  quality 
down  to  93.15  per  cent  or  6.85  per  cent  of  priming.  The  range  of  the  apparatus  is 
evidently  extended  by  connecting  the  outlet  to  the  condenser  and  thus  using  a  lower 
back  pressure. 

52.  Throttle  control.     A  small  non-condensing  engine  uses  steam  at 
50  pounds  absolute,  receiving  it  through  a  throttle  valve  from  a  header 
containing  dry  steam  at  200  pounds  absolute.1     The  atmospheric 
pressure  is  14.7  pounds.     Find  the  loss  in  availability  of  the  steam 
in  B.t.u.  per  pound  due  to  throttling. 

Solution.  During  throttling  heat  is  converted  into  kinetic  energy  and  then  by  fric- 
tion into  heat  as  in  the  calorimeter  of  Problem  50.  The  0-0  diagram  for  throttling  is 
similar  to  that  of  Problem  50. 

If  a  quantity  of  heat  dO  is  received  from  a  source  at  a  temperature  of  0,  and  if  the 
lowest  available  temperature  is  00,  the  Carnot  efficiency  between  these  two  temperatures 
represents  the  fraction  of  dO  which  is  available  for  useful  work.  This  available  porT 
tion  of  cZCl,  the  heat  received  from  the  hot  source,  has  been  termed  its  "  motivity  "  by 
Lord  Kelvin.2  Now  the  heat  possessed  by  a  mass  of  steam  has  been  received  at  a 
series  of  temperatures  ranging  in  the  boiler  from  that  of  the  feed  water  as  it  enters  to 
its  final  temperature.  At  each  of  these  temperatures  the  motivity  or  availability  is 

given  by  an  expression  of  the  form  £xfe  dO.    The  availability  of  a  given  mass  in  any 

6 

state,  as  1,  is  then  the  sum  of  all  the  motivities  due  to  the  addition  of  the  several  infin- 
itesimal amounts  of  heat  dO  at  these  various  temperatures.  That  is,  the  motivity  in 
state  1  is  r 

\    6-   =^0  =  01-0001. 

Jo      * 

Similarly,  the  motivity  in  any  other  state,  as  2,  is  of  the  form  02  —  0002-  The  differ- 
ence in  availability  of  the  substance  at  these  two  states  is  then 

Oi-O2  +  00(>2-0i), 

1  The  conditions  here  chosen  are  not  such  as  would  ordinarily  enter  into  good  engineer- 
ing practice.    They  are  selected  so  that  02  — 0i  is  large  as  compared  to  the  precision  with 
which  0J  or  02  is  known. 

2  See  Preston,  "  Heat,"  pp.  629,  630. 


122  THEKMODYNAMICS 

an  expression  which  for  the  condition  of  constant  heat  contents  required  by  this  par- 
ticular problem  reduces  to 


The  lowest  temperature  for  this  problem  is  that  at  which  the  steam  is  rejected,  or 
212°  Fahrenheit.  The  absolute  temperature  00  corresponding  to  this  is  672.  The  en- 
tropy 0i  is  found  directly  from  Table  3  to  be  1.5459. 

In  order  to  find  the  entropy  in  state  2  it  is  necessary  to  make  use  of  the  condition 
that  the  heat  contents  remain  constant  and  solve  for  the  temperature  as  follows  : 

*i  =  01  +  n  =  354.3  +  843.5  =  1197.8, 

X2  =  £2  +  rz  +  cp(t2  -  t^  =  250.4  +  922.8  +  cp(t2  -  281)  =  1173.2  +  h. 
Therefore 

ft  =  24.6;  £2-«0=51.2.     0g=  -  ^  -  =  0.0324. 

460  +  $  (281  +332) 

The  entropy  in  the  saturated  condition  is  found  from  the  tables,  hence  the  total 
entropy  02  is  1.6581  +  0.0324  =  1.6905.  The  loss  in  availability  per  pound  is  then 

672(1.6905  -  1.5459)  =  97.2  B.  t.  u. 

53.  Nozzle  area.     Plot  to  a  scale  of  pressures  for  abscissas  and 
ordinates  of  B.  t.  u.  the  energy  \  —  X  available  for  increasing  the 
kinetic  energy  of  a  mixture  of  steam  expanding  adiabatically  with- 
out friction  from  150  pounds  absolute  and  75°  Fahrenheit  of  super- 
heat to  a  back  pressure  of  6  pounds  absolute.     Plot  also  the  velocity 
during  this  expansion.     Plot  the  area  in  square  inches  of  a  nozzle 
to  discharge  one  pound  per  second. 

Solution.  From  Table  3,  for  150  pounds,  gi  =  330.0,  n  =  863.0.  For  75  degrees 
of  superheat,  use  cp  =  0.57.  Hence  hi  =  43.  Therefore  Xi  =  1236.  0Wl  =  0.5142, 
<f>Vi  =  1.0551,  and  0,t  =  0.0503.  Hence  0j.  =  1.6196  or  1.62  approximately. 

The  values  of  the  heat  contents  for  other  pressures  may  be  found  either  by  calcula- 
tion or  they  may  be  read  directly  from  the  pressure-entropy  chart  of  page  86.  In  this 
solution  they  are  read  directly  from  the  chart,1  and  also  the  values  of  the  specific  vol- 
umes corresponding.  The  velocity  is  then  found  by  equation  (71)  and  the  requisite 
area  by  equation  (65),  which  may  be  written 

(144)  (specific  volume) 

area  m  square  inches  =  -  -  ^      .  —  r—      —  -  • 

velocity 

The  desired  values  are  plotted  in  Figure  64. 

54.  Turbine  nozzle.     A  300  horse-power   De  Laval  turbine  uses 
18  pounds  of  steam  per  horse-power  hour.     There  are  twelve  nozzles 
supplying  steam  to  the  turbine  wheel.     The  steam  is  at  150  pounds 
absolute  and  75  degrees  of  superheat,  and  it  expands  to  a  back  press- 
ure of  2  pounds  absolute.     Find  the  diameters  of  the  nozzle  at  the 
throat  and  at  the  exit  end  necessary  to  deliver  the  requisite  amount 

1  For  the  purposes  of  this  calculation  the  diagram  proposed  by  Professor  Mollier  and 
named  after  him  is  most  convenient.  In  it  ordinates  are  X's  and  abscissas  are  0's.  Lines 
of  constant  pressure  and  of  quality  are  plotted.  Such  a  diagram  is  contained  in  the  Tables 
of  Marks  and  Davis. 


FLOW   OF   FLUIDS 


123 


of  steam  :   (#)  for  frictionless  adiabatic  expansion,  (5)  for  15  per  cent 
energy  loss  by  friction.     For  the  throat  pressure 


3OOO 


250 


2000    200 


150 


1POO     100 


50 


7* 


2.25 
2.00 
1.75 
1.50 
1.25 
1.OO 
0.75 
0.50 
O.25 


150  140 


120 


60 


Pounds 
FIG.  64 


Solution,     (a)  Each  nozzle  must  deliver    (  t    ^ — ^- ,  or  0.125  pound  per  second. 

(3600) (12) 

The  quality  in  the  throat  for  adiabatic  expansion  is  found  as  follows.  The  entropy 
is  1.6196  in  the  initial  condition  as  found  in  Problem  53.  For  the  throat  pressure 
(0jo)f  =  0.4617  and  (0y),  —  1.1516.  Hence  (0S),  must  equal  0.0063.  Since  the  satura- 
tion temperature  is  318°  Fahrenheit,  for  the  throat  pressure  of  87  pounds,  the  quality 
is  10  degrees  of  superheat  (using  cp  =  0.48). 
By  equation  (57)  the  specific  volume  is 

(86.86)  (788)  _  0.256  =  5.13. 

(87) (144) 

The  total  heat  content  at  the  throat  is  (\)t  -  288.2  +  895.3  +  4.8  =  1188.3. 
The  total  heat  contents  in  the  initial  condition  were  found  in   Problem  53   to  be 
\±  =  1236. 

The  velocity  in  the  throat  is  then 


b2  =  V2  g  778  (1236  —  1188.3)  =  1545  feet  per  second. 
The  requisite  area  in  the  throat  must  then  be 

i ^ ' '  '  '  —  0.479  square  inch  per  pound  of  steam. 

But  for  the  turbine  nozzle  of  the  problem  only  0.125  pounds  per  second  are  to  pass, 
hence  the  requisite  area  is  (0. 125)(0.479)  =  0.06  square  inches.  Ihe  diameter  at  the 
throat  is  0.276  inch. 


124  THERMODYNAMICS 

Similar  solution  for  the  exit  end  gives  x2  =  0.828,  X2  =  940.0,  and  the  specific  volume 
at  the  exit  is  143.3  cubic  feet.  The  velocity  is  therefore  3835  feet  per  second,  the  area 
of  the  turbine  nozzle  is  0.672  square  inch,  and  the  diameter  is  0.924  inch. 

(&)  It  is  usual  not  to  make  allowance  for  the  friction  at  the  throat,  since  its  effect 
would  be  small,  but  to  make  the  nozzle  at  the  throat  according  to  the  calculations 
for  a  frictionless  adiabatic  expansion  as  made  above.  It  will  be  remembered  in  justi- 
fication of  this  method  that  the  steam  discharged  through  an  orifice  is  practically 
independent  of  the  pressures,  provided  only  that  the  exit  pressure  is  less  than  0.58  of 
the  initial  pressure.  (Thus  compare  Problem  47.) 

At  the  exit  end  the  loss  in  energy  is  15  per  cent  of  Xi  —  X2  or  44  B.t.  u.  The 
quality  at  exit  for  a  flow  with  friction  is  greater  than  that  for  a  frictionless  flow  by  an 
amount  equal  to  the  fraction  of  a  pound  of  steam  that  this  number  of  B.  t.  u.  would 

evaporate.     Since  r2  for  2  pounds  is  1021.9,  the  increase  in  quality  is  -  =  0.043.    The 

1022 

quality  after  flow  with  friction  is  then  0.828  +  0.043  =  0.871.  Hence  the  specific 
volume  is  greater  and  is  (0.871)  (173.1)  =  150.8  cubic  feet.  In  order  that  this  turbine 
nozzle  shall  deliver  the  same  amount  of  steam  for  the  same  pressure  conditions  the 
area  at  the  exit  end  must  be  larger  than  that  for  the  case  of  frictionless  expansion  in 
the  ratio  of  the  specific  volumes  in  the  two  cases.  Hence  the  requisite  area  is 

(0.672)  =0.706  square  inch. 


143.3 
And  the  diameter  is  0.947  inch. 


MISCELLANEOUS   PROBLEMS 

1.  Determine  four  points  on  the  pressure-volume  curve  and  also 
four  points  on  the  temperature-entropy  curve  representing  the  trans- 
formation of  a  pound  of  dry  air  undergoing  an  increase  of  tempera- 
ture from  100°  to  200°  Fahrenheit  under  the  following  conditions : 
(a)  adiabatically  from  an  initial  pressure  of  15  pounds,  (5)  at  con- 
stant pressure  of  15 l  pounds  per  square  inch,  (<?)  at  constant  volume 
of  11  cubic  feet.     Assume  a  zero  of  entropy  for  each  initial  condition. 

2.  Plot  in  the  same  manner  as  in  Problem  1  the  transformation 
at  a  constant  temperature  of  one  pound  of  air  originally  occupying 
11  cubic  feet  at  35  pounds  per  square  inch  if  the  pressure  decreases 
to  15  pounds. 

3.  Find  the  work  done,  the  change  in  internal  energy,  the  change 
in  entropy,  and  the  heat  added  during  the  expansion  of  3.5  cubic 
feet  of  air  from  200  to  90  pounds  pressure.     The  temperature  is 
maintained  constant  at  650°  Fahrenheit  absolute. 

4.  Find  the  external  work  done,  the  change  in  internal  energy, 
the  change  in  entropy,  and  the  heat  added  if  3  cubic  feet  of  air  at 
200°  Fahrenheit  expand  adiabatically  from  a  pressure  of  85  pounds 
per  square  inch  to  a  pressure  of  15  pounds. 

5.  A  quantity  of  gas  occupying  9  cubic  feet  at  a  pressure  of  42 
pounds  per  square  inch  expands  until  the  volume  is  14  cubic  feet  at 
a  pressure  of  23  pounds.     Find  an  equation  representing  a  possible 
expansion  between  these  two  points.     Find  the  work  done.     Find 
the  change  in  intrinsic  energy. 

6.  A  steam  engine  working  between  135  pounds  and  2  pounds 
per  square  inch  pressure  requires  225  B.  t.  u.  per  indicated  horse 
power  per  minute.     What  is  the  ratio  between  the  thermal  efficiency 
of  this  engine  and  that  of  a  Carnot  engine  working  between  the 
same  temperature  limits  ? 

7.  Calculate  the  work  done  by  an  air  compressor  without  clear- 
ance in  compressing  and  delivering  10  cubic  feet  of  dry  air  from  an 
atmospheric   pressure   of    15   pounds   to   a   gauge    pressure   of   75 
pounds.     Assume  adiabatic  compression. 

1  All  pressures  are  absolute  and  not  gauge  unless  otherwise  stated. 

125 


126  THERMODYNAMICS 

8.  What  would  have  been  the  work  in  Problem  7  if  the  com- 
pression had  been  according  to  the  relation  p  V1'^  —  a  constant  ? 

9.  If  cooling  water  of  temperature  65°  Fahrenheit  is  available,  to 
what  pressure  must  dry  atmospheric  air  be  compressed  in  order  that 
after  cooling  it  may  in  an  adiabatic  expansion  fall  to  32°  Fahrenheit  ? 
The  initial  temperature  of  the  air  is  70°  Fahrenheit.     Atmospheric 
pressure  is  15  pounds  per  square  inch. 

10.  Find  from  the  entropy  chart  of  page  86  what  the  pressure  is 
upon  a  mixture  of  steam  and  water  one  pound  of  which  occupies 
6  cubic  feet,  if  the  quality  is  95  per  cent. 

11.  What   is  the   quality  of   a   mixture  of   steam  and  water  of 
entropy  1.590  if  one  pound ,  occupies  8  cubic  feet? 

12.  What  is  the  quality  in  Problem  11  if  the  mixture  occupies 
3  cubic  feet  ? 

13.  One  pound  of  saturated  steam  at  a  pressure  of  140  pounds 
per  square  inch  undergoes  a  constant  heat  content  change  until  the 
final  pressure  is  25  pounds.     Find  the  quality.     Calculate  the  final 
intrinsic  energy. 

14.  Steam  at  140  pounds  gauge  and  superheated  100°  Fahrenheit 
expands  adiabatically.     What  is  the  pressure  when  the  steam  is  dry 
saturated  steam  ?     Atmospheric  pressure  is  15  pounds. 

15.  If  the  steam  of  Problem  14  expands  in  an  engine  cylinder  to 
five   times  its  original  volume,  what,  approximately,  are  the   final 
pressure  and  quality? 

16.  Steam    of    entropy   1.560   and    heat    contents   1200   B.t.u. 
expands  adiabatically  until   200  B.  t.  u.  have   been   converted  into 
external  work.     Find  the  final  quality. 

17.  Find  from  the  entropy  chart  of  page  86  the  limit  of  a  throt- 
tling calorimeter  when  used  to  measure  quality  in  steam  of  pressure 
145  pounds  if  it  is  connected  to  the  atmosphere  of  which  the  absolute 
pressure  is  14.7  pounds. 

18.  Steam  of  quality  99  per  cent  and  pressure  160  pounds  passes 
through  a  reducing  valve.     To  what  pressure  must  it  be  throttled 
in  order  that  the  steam  discharged  shall  be  dry  and  saturated? 

19.  Steam  from  a   boiler  is  passed  into   a  barrel  calorimeter  in 
which  there  is  200  pounds  of  water  at  a  temperature  of  60°  Fahren- 
heit.    When  3  pounds  have  been  condensed,  the  final  temperature  is 
75°   Fahrenheit.     The   boiler    pressure   is   145   pounds.     Find   the 
quality. 


MISCELLANEOUS   PROBLEMS  127 

20.  Feed  water  of  temperature  80°  Fahrenheit  enters  a  boiler  and 
passes   from  a  boiler   through   a   superheater.     It   emerges  at  100 
pounds   pressure   and   a    temperature    of    600°   Fahrenheit.       If   8 
pounds  of  water  is  thus  converted  into  steam  for  each  pound  of  coal 
burned,  find  the  equivalent  evaporation  from  and  at  212°  Fahren- 
heit. 

21.  Find  from  values  taken  from  the  entropy  chart  of  page  86 
the  Rankine  efficiency  of  an  engine  working  between  140  pounds 
with  a  quality  of  95  per  cent  and  10  pounds. 

22.  A  four-stage  impulse  turbine  is  designed  to  develop  equal 
velocity  in   each   of   the  stages.     If   supplied  with   steam   of   100 
pounds  pressure  and  100°  Fahrenheit  superheat  and  expanding  to 
a  condenser  pressure  of  2   pounds,  find  the  velocity  developed  in 
each  stage. 

23.  What  is  the  area  at  the  throat  of  a  nozzle  which  allows  50 
pounds  of  steam  to  pass  in  a  minute,  if  the  entering  steam  is  150 
pounds  pressure  and  100°  Fahrenheit  superheat,  and  the  back  press- 
ure is  less  than  0.58  of  the  admission  pressure  ? 

24.  What  would  be  the  area  if  the  back  pressure  in  Problem  23 
were  100  pounds? 

25.  How  much  work  is  done  in  compressing   adiabatically  one 
pound  of  a  mixture  of  steam  and  water  of  quality  80  per  cent  and 
pressure  14.7  pounds  until  the  volume  is  6  cubic  feet? 

26.  Two  pipes  deliver  into  a  third.     One  supplies  300  gallons  per 
minute  at  a  temperature  of  80°  Fahrenheit  and  the  other  supplies 
steam  of  quality  75  per  cent  and  pressure  100  pounds  per  square  inch 
at  the  rate  of  5  pounds  per  minute.     What  is  the  resulting  quality 
of  the  mixture  in  the  third  pipe  ? 

27.  A  hot-water   heater  is  constructed   on   the  above  principle 
(Problem  26).     Ten  gallons  of  water  at  200°  are  to  be  obtained  per 
minute  in  the  third  pipe.     The  entering  water  is  at  60°  Fahrenheit. 
The  entering  steam  is  at  50  pounds  pressure  and  dry.     What  is  the 
smallest  amount  of  steam  in  pounds  required  per  minute  ? 

28.  What  approximately  is  the  final  temperature  and  pressure 
exerted  by  a  mixture  of  steam  and  water  in  a  closed  tank  if  the  tank 
originally  contained  10  pounds   at  a  pressure   of  140   pounds  per 
square  inch  and  a  quality  of  98  per  cent  and  9000  B.  t.u.  are  removed 
by  cooling? 


128  THERMODYNAMICS 

29.  Find  the  increase  in  volume  per  pound  caused  by  passing  dry 
steam  at  a  pressure  of  130  pounds  per  square  inch  through  a  super- 
heater which  raises  the  temperature  to  500°  Fahrenheit. 

30.  A   throttle  valve   reduces   the  pressure  of  dry  steam  from 
135  pounds  to  100  pounds.     Find  the  loss  in  availability  in  B.t.  u. 
per  pound  if  the  lowest  available  temperature  is  141.5°  Fahrenheit. 

31.  Plot  to  scale  the  indicator  card  for  a  theoretically  perfect 
engine  which  has.  a  stroke  of  30  inches,  a  clearance  of  5  per  cent, 
cut-off  at  30  per  cent,  compression  at  5  per  cent,  release  at  the  end  of 
the  stroke,  a  piston  diameter  of  16  inches,  and  makes  100  revolutions 
per  minute.     If   the  entering  steam  is   100  pounds  pressure,  the 
condenser  pressure  is  2  pounds,  and  the  expansion  and  compression 
may  be  assumed  to  be  adiabatic,  find  the  work  done  during  admission, 
expansion,  exhaust,    and   compression.     Find    the    indicated   horse 
power.     Find  the  mean  effective  pressure.     The  engine  is  of  course 
double-acting,  having  valves  at  each  end  of  the  cylinder. 

32.  Find  the  horse  power  of  a  hot-air  engine  which  works  on  a 
Sterling  cycle  from  a  maximum  temperature  of  700°  Fahrenheit  and 
a  maximum  pressure  of  110  pounds  absolute,  if  the  volume  of  the 
working  cylinder  is  2  cubic  feet  and  that  of  the  displacement  cylinder 
is   6.5   cubic  feet.     Neglect  clearance.     The  lowest  available  tem- 
perature is  90°  Fahrenheit.     The  engine  makes  100  revolutions  per 
minute. 

33.  Draw  to  scale  the  cross  section  of  a  nozzle  which  will  discharge 
0.50  pound  of  steam  per  second  for  frictionless  adiabatic  expansion 
from  120  pounds  pressure  and  120°  Fahrenheit  superheat  to  2  pounds 
pressure  per  square  inch. 

34.  If  the  friction  loss  in  the  nozzle  of  Problem  33  is  to  be  taken 
as  12  per  cent  and  the  nozzle  is  drawn  tapered  from  the  throat  to  the 
exit  end,  how  much  larger  must  be  the  diameter  of  the  exit  end  in 
order  that  the  same  amount  of  steam  shall  pass  per  second  ? 

35.  If  the  steam  of  Problem  33  had  not  been  superheated  but  had 
been  of  quality  98  per  cent,  what  would  have  been  the  velocity  at 
the  exit  end  of  the  nozzle  ? 

36.  If  a  steam  turbine  is  to  have  four  stages  of  equal  velocity  and 
to  be  supplied  with  steam  under  the  conditions  of  Problem  33,  what 
will  be  the  pressures  in  the  intermediate  stages  ? 


TABLES 


129 


TABLES 

TABLE   1 


GAS 

EXPANSION 
COEFFICIENT  a 

PRESSURE 
COEFFICIENT  ft 

OBSERVER 

DATE 

Hydrogen    
Hydrogen 

0.0036600 
0  003661 

0.0036626 
0  003668 

P.  Chappuis 

Regnault 

1887 
1840 

Air 

0  003670 

0  003665 

Rggnault 

1840 

Air      
Carbon  dioxide  (C02)     . 
Nitrous  oxide  (N20)       . 
Oxvsren 

0.003710 
0.003719 

0.003663 
0.003688 
0.003676 
0  003674 

Kuenen  and  Randall 
Regnault 
Regnault 

1895 
1840 
1840 

1874 

Nitrogen 

0  003667 

von  Jolly 

1874 

Argon 

0  003668 

Kuenen  and  Randall 

1895 

Helium    

0.0036627 

Travers  and  Jaquerod 

1903 

TABLE  2 


SUBSTANCE 

SPECIFIC  VOLUME1 
Cubic  feet  per  pound 

SPECIFIC  HEAT 
Constant  pressure 

SPECIFIC  HEAT 
Constant  Volume 

KATIO 
*-2 

Cv 

Hydrogen      .     .     . 

178.2 

3.409 

2.406 

1.408 

Oxygen     .... 

12.21 

0.2175 

0.155 

1.403 

Air  (dry)  .... 

12.39 

0.2375 

0.1689 

1.401 

Nitrogen  .... 

12.75 

0.2438 

0.  173 

1.409 

Marsh  gas  (CH4)  . 

23.30 

0.593 

0.467 

1.31 

Carbon  dioxide 

8.10 

0.2025 

0.171 

1.30 

Carbon  monoxide  . 

12.80 

0.2426 

0.173 

1.401 

Ethlyene  (C2H4)    . 

12.58 

0.404 

0.332 

1.24 

1  For  standard  conditions  of  p  =  14.7  pounds  per  square  inch  and  t  =  32°  Fahrenheit. 


130 


THERMODYNAMICS 


131 


TABLE  31 

(F.P.  S.  Gravitational  system;  degrees  Fahrenheit,  British  thermal  units) 

DRY  SATURATED  STEAM  , 


a 

11 

H 

I 

• 

H 

H 

j 

H 

-*;   -- 

<  < 

g 

a  H 

H 

o 

|« 

£| 

1 

>o 

1 

£>  H 

H 
O 

w  o 

PH    O 

§ 

»i 

fc 

II 

I  » 

°  5 

0  § 

6 

w  £  a 

0-H 

o 

&ls 

£  8 

£  § 

-I 

Bj 

H  <y 

H   S 

tt 

Ei1""1  0 

E«"  O 

i 

•   s  *=  M 

§  2" 

§  % 

go 

as  M 

W  ^ 

w  o 

» 

W  O^ 

H  Op* 

w 

figP-lO 

fc^ 

65^*" 

c  W 

EH 

PH  ^ 

W 

H 

H 

a 

£ 

OD 

Q 

m 

w 

H 

t 

* 

« 

f 

A 

p 

Apu 

8 

i 

8 

*. 

* 

*.+*. 

60 

0.256 

28.1 

1057.0 

1085.1 

999.8 

57.2 

1207. 

0.000828 

0.0556 

2.0347 

2.0903 

65 

0.305 

33.1 

1054.4 

1087.5 

996.7 

57.7 

1021. 

0.000979 

0.0652 

2.0103 

2.0755 

70 

0.363 

38.1 

1051.8 

1089.9 

993.6 

58.2 

868. 

0.001152 

0.0747 

1.9863 

2.0610 

75 

0.429 

43.1 

1049.2 

1092.3 

990.5 

58.7 

739. 

0.001353 

0.0841 

1.9629 

2.0470 

80 

0.506 

48.1 

1046.5 

1094.6 

987.2 

59.3 

634. 

0.001577 

0.0934 

1.9398 

2.0332 

85 

0.594 

53.1 

1043.9 

1097.0 

984.1 

59.8 

544. 

0.001838 

0.1026 

1.9171 

2.0197 

90 

0.696 

58.1 

1041.2 

1099.3 

980.9 

60.3 

469. 

0.002131 

0.1117 

1.8948 

2.0065 

100 

0.946 

68.0 

1035.7 

1103.7 

974.4 

61.3 

351.0 

0.002851 

0.1297 

1.8611    .9808 

101.8 

1 

69.8 

1034.7 

1104.5 

973.1 

61.6 

333.0 

0.00300 

0.1329 

1.8433    .9762 

110 

1.271 

78.0 

1030.1 

1108.1 

967.7 

62.4 

265.0 

0.003771 

0.1473 

1.8088    .9561 

120 

1.689 

88.0 

1024.4 

1112.4 

961.0 

63.4 

203.0 

0.004926 

0.1647 

1.7677 

.9324 

126.1 

2 

94.2 

1021.9 

1116.1 

957.8 

64.1 

173.0 

0.00578 

0.1753 

1.7432 

.9185 

141.5 

3 

109.6 

1012.2 

1121.8 

946.4 

65.8 

118.4 

0.00845 

0.2011 

1.6841 

.8852 

153.0 

4 

121.0 

1005.5 

1126.5 

938.6 

66.9 

90.4 

0.01106 

0-2200 

1.6416 

.8616 

162.3 

5 

130.3 

1000.0 

1130.3 

932.1 

67.9 

73.3 

0.01364 

0.2351 

1.6084 

1.8435 

170.1 

6 

138.1 

995.5 

1133.6 

926.8 

68.7 

61.9 

0.01616 

0.2476 

1.5812 

1.8288 

193.2 

10 

161.3 

981.4 

1142.7 

910.4 

71.0 

38.37 

0.02606 

0.2838 

1.5036 

1.7874 

212 

14.7 

180.3 

969.7 

1150.0 

896.9 

72.8 

26.78 

0.03734 

0.3125 

1.4441 

1.7566 

228.0 

20 

196.4 

959.4 

1155.8 

885.1 

74.3 

20.09 

0.04978 

0.3362 

1.3957 

1.7319 

250.3 

30 

219.1 

944.4 

1163.5 

868.2 

76.2 

13.74 

0.0728 

0.3687 

1.3305 

1.6992 

281.0 

50 

250.4 

922.8 

1173.2 

844.1 

78.7 

8.507 

0.1176 

0.4119 

1.2462 

1.6581 

303.0 

70 

272.9 

906.6 

1179.5 

826.3 

80.3 

6.199 

0.1613 

0.4418 

1.1892 

1.6310 

312.1 

80 

282.2 

899.8 

1182.0 

818.9 

80.9 

6.466 

0.1829 

0.4540 

1.1661 

1.6201 

316.3 

85 

286.5 

896.6 

1183.1 

815.4 

81.2 

5.161 

0.1938 

0.4595 

1.1557 

1.6152 

320.3 

90 

290.7 

893.5 

1184.2 

812.1 

81.4 

4.886 

0.2047 

0.4649 

1.1457 

1.6106 

324.2 

95 

294.6 

890.5 

1185.1 

808.8 

81.7 

4.644 

0.2153 

0.4699 

1.1363 

1.6062 

327.9 

100 

298.5 

887.6 

1186.1 

805.7 

81.9 

4.432 

0.2256 

0.4748 

1.1273 

1.6021 

331.4 

105 

302.1 

884.8 

1186.9 

802.7 

82.1 

4.233 

0.2362 

0.4794 

1.1187 

1.5981 

334.8 

110 

305.6 

882.1 

1187.7 

799.7 

82.4 

4.047 

0.2471 

0.4838 

1.1105 

1.5943 

338.1 

115 

309.0 

879.5 

1188.5 

797.0 

82.5 

3.876 

0.2580 

0.4881 

1.1026 

1.5907 

341.3 

120 

312.3 

876.9 

1189.2 

794.2 

82.7 

3.723 

0.2686 

0.4922 

1.0951 

1.5873 

344.4 

125 

315.5 

874.5 

1190.0 

791.6 

82.9 

3.581 

0.2793 

0.4962 

1.0878 

1.5840 

347.4 

130 

318.6 

872.1 

1190.7 

789.0 

83.1 

3.451 

0.2898 

0.5000 

1.0808 

1.5808 

350.3 

135 

321.5 

869.8 

1191.3 

786.5 

83.3 

3.331 

0.3002 

0.5037 

1.0741 

1.5778 

353.1 

140 

324.4 

867.4 

1191.8 

784.0 

83.4 

3.220 

0.3106 

0.5073 

1.0675 

1.5748 

355.8 

145 

327.3 

865.2 

1192.5 

781.6 

83.6 

3.115 

0.3210 

0.5108 

1.0612 

1.5720 

358.5 

150 

330.0 

863.0 

1193.0 

779.3 

83.7 

3.014 

0.3318 

0.5142 

1.0551 

1.5693 

363.6 

160 

335.3 

858.8 

1194.1 

774.9 

83.9 

2.834 

0.3528 

0.5206 

1.0434 

1.5640 

368.5 

170 

340.4 

854.8 

1195.2 

770.6 

84.2 

2.673 

0.3741 

0.5268 

1.0324 

1.5592 

373.2 

180 

345.2 

850.9 

1196.1 

766.5 

84.4 

2.531 

0.3951 

0.5326 

1.0219 

1.5545 

381.9 

200 

354-3 

843.5 

1197.8 

758.8 

84.7 

2.288 

0.4371 

0.5434 

1.0025 

1.5459 

401.1 

250 

374.2 

826.9 

1201.1 

741.5 

85.4 

1.845 

0.542 

0.5669 

0.9609 

1.5278 

1  These  values  are  taken  from  the  1909  edition  of  Professor  C.  H.  Peabody's  "  Tables  of  the 
Properties  of  Steam,"  with  the  permission  of  the  author  and  the  publishers,  Wiley  and  Sons. 


132 


THERMODYNAMICS 


LOGARITHMS 


10 

0 

l 

2 

3 

4 

5 

6 

7 

8 

9 

123 

456 

789 

0000 

0043 

0086 

0128 

0170 

0212 

0253 

0294 

0334 

0374 

4  8  12 

17  21  25 

29  33  37 

11 

12 
13 

0414 
0792 
1139 

0453 
0828 
1173 

0492 
0864 
1206 

0531 
0899 
1239 

0569 
0934 
1271 

0607 
0969 
1303 

0645 
1004 
1335 

0682 
1038 
1367 

0719 
1072 
1399 

0755 
1106 
1430 

4  8  11 
3  7  10 
3  6  10 

15  19  23 
14  17  21 
13  16  19 

26  30  34 

24  28  31 
23  26  29 

14 
15 
16 

1461 
1761 
2041 

1492 
1790 
2068 

1523 

1818 
2095 

1553 
1847 
2122 

1584 
1875 

2148 

1614 
1903 
2175 

1644 
1931 
2201 

1673 
1959 

2227 

1703 
1987 
2253 

1732 
2014 
2279 

369 
368 

358 

12  15  18 
11  14  17 
11  13  16 

21  24  27 
20  22  25 
18  21  24 

17 

18 
19 

2304 
2553 

2788 

2330 

2577 
2810 

2355 
2601 
2833 

2380 
2625 

2856 

2405 

2648 

2878 

2430 
2672 
2900 

2455 
2695 
2923 

2480 
2718 
2945 

2504 

2742 
2967 

2529 
2765 
2989 

257 
257 

24  7 

10  12  15 
9  12  14 
9  11  13 

17  20  22 
16  19  21 
16  18  20 

20 

3010 

3032 

3054 

3075 

3096 

3118 

3139 

3160 

3181 

3201 

24  6 

8  11  13 

15  17  19 

21 
22 
23 

3222 
3424 
3617 

3243 
3444 
3636 

3263 
3464 
3655 

3284 
3483 
3674 

3304 
3502 
3692 

3324 
3522 
3711 

3345 
3541 
3729 

3365 
3560 

3747 

3385 
3579 
3766 

3404 
3598 

3784 

246 
246 
246 

8  10  12 
8  10  12 
7  0  11 

14  16  18 
14  15  17 
13  15  17 

24 
25 
26 

3802 
3979 
4150 

3820 
3997 
4166 

3838 
4014 
4183 

3856 
4031 
4200 

3874 
4048 
4216 

3892 
4065 
4232 

3909 

4082 
4249 

3927 
4099 
4265 

3945 
4116 

4281 

3962 
4133 
4298 

245 
235 

235 

7  0  11 
7  9  10 

7  8  10 

12  14  16 
12  14  15 
11  13  15 

27 

28 
29 

4314 

4472 
4624 

4330 
4487 
4639 

4346 
4502 
4654 

4362 

4518 
4669 

4378 
4533 
4683 

4393 

4548 
4698 

4409 
4564 
4713 

4425 
4579 

4728 

4440 
4594 

4742 

4456 
4609 

4757 

235 
235 
1  3  4 

689 
689 
679 

11  13  14 
11  12  14 
10  12  13 

30 

4771 

4786 

4800 

4814 

4829 

4843 

4857 

4871 

4886 

4900 

1  3  4 

679 

10  11  13 

31 
32 
33 

4914 
5051 
5185 

4928 
5065 
5198 

4942 
5079 
5211 

4955 
5092 
5224 

4969 
5105 
5237 

4983 
5119 
5250 

4997 
5132 
5263 

5011 
5145 
5276 

5024 
5159 
5289 

5038 
5172 
5302 

1  3  4 
1  3  4 
1  3  4 

678 
578 
568 

10  11  12 
9  11  12 
9  10  12 

34 
35 
36 

5315 
5441 
5563 

5328 
5453 
5575 

5340 
5465 

5587 

5353 

5478 
5599 

5366 
5490 
5611 

5378 
5502 
5623 

5391 
5514 
5635 

5403 
5527 
5647 

5416 
5539 
5658 

5428 
5551 
5670 

1  3  4 
1  2  4 
1  2  4 

568 
567 
567 

9  10  11 
9  10  11 
8  10  11 

37 
38 
39 

5682 
5798 
5911 

5694 
5809 
5922 

5705 
5821 
5933 

5717 
5832 
5944 

5729 
5843 
5955 

5740 
5855 
5966 

5752 
5866 
5977 

5763 

5877 
5988 

5775 
5888 
5999 

5786 
5899 
6010 

1  2  3 
133 
1  2  3 

567 

567 

457 

8  9  10 
8  9  10 
8  9  10 

40 

6021 

6031 

6042 

6053 

6064 

6075 

6085 

6096 

6107 

6117 

1  2  3 

456 

8  9  10 

41 
42 
43 

6128 
6232 
6335 

6138 
6243 
6345 

6149 
6253 
6355 

6160 
6263 
6365 

6170 
6274 
6375 

6180 
6284 
6385 

6191 
6294 
6395 

6201 
6304 
6405 

6212 
6314 
6415 

6222 
6325 
6425 

123 
1  2  3 
123 

456 
456 

456 

789 
789 
789 

44 
45 
46 

6435 
6532 
6628 

6444 
6542 
6637 

6454 
6551 
6646 

6464 
6561 
6656 

6474 
6571 
6665 

6484 
6580 
6675 

6493 
6590 
6684 

6503 
6599 
6693 

6513 
6609 
6702 

6522 
6618 
6712 

123 
123 
1  2  3 

456 
456 
456 

789 
789 

778 

47 

48 
49 

6721 
6812 
6902 

6730 
6821 
6911 

6739 
6830 
6920 

6749 
6839 
6928 

6758 
6848 
6937 

6767 
6857 
6946 

6776 
6866 
6955 

6785 
6875 
6964 

6794 
6884 
6972 

6803 
6893 
6981 

1  2  3 
1  2  3 
1  2  3 

455 
445 
445 

678 
678 
678 

50 

6990 

6998 

7007 

7016 

7024 

7033 

7042 

7050 

7059 

7067 

1  2  3 

345 

678 

51 
52 
63 

7076 
7160 
7243 

7084 
7168 
7251 

7093 

7177 
7259 

7101 
7185 
7267 

7110 
7193 

7275 

7118 
7202 
7284 

7126 
7210 
7292 

7135 

7218 
7300 

7143 
7226 

7308 

7152 
7235 
7316 

123 
1  2  2 
1  2  2 

345 
345 
345 

678 
677 
667 

54 

7324 

7332 

7340 

7348 

7356 

7364 

7372 

7380 

7388 

7396 

122 

345 

667 

TABLES 


133 


LOGARITHMS 


55 

0 

1 

2 

3 

4 

5 

6 

7 

8 

9 

1  2  3 

456 

789 

7404 

7412 

7419 

7427 

7435 

7443 

7451 

7459 

7466 

7474 

122 

345 

567 

56 
57 

58 

7482 
7559 
7634 

7490 
7566 
7642 

7497 
7574 
7649 

7505 
7582 
7657 

7513 
7589 
7664 

7520 
7597 
7672 

7528 
7604 
7679 

7536 
7612 

7686 

7543 
7619 
7694 

7551 
7627 
7701 

122 
1  2  2 
112 

345 
345 
344 

567 
567 
567 

69 
60 
61 

7709 
7782 
7853 

7716 

7789 
7860 

7723 
7796 

7868 

7731 

7803 

7875 

7738 
7810 
7882 

7745 
7818 
7889 

7752 
7825 
7896 

7760 
7832 
7903 

7767 
7839 
7910 

7774 
7846 
7917 

112 
112 
112 

344 
344 

344 

567 
566 
566 

62 
63 
64 

7924 
7993 
8062 

7931 
8000 
8069 

7938 
8007 
8075 

7945 
8014 
8082 

7952 
8021 
8089 

7959 
8028 
8096 

7966 
8035 
8102 

7973 
8041 
8109 

7980 
8048 
8116 

7987 
8055 
8122 

1  1 
1  1 
1  1 

334 
334 
334 

566 
556 
556 

65 

8129 

8136 

8142 

8149 

8156 

8162 

8169 

8176 

8182 

8189 

1  1 

334 

556 

66 
67 

68 

8195 
8261 
8325 

8202 
8267 
8331 

8209 

8274 
8338 

8215 
8280 
8344 

8222 

8287 
8351 

8228 
8293 
8357 

8235 
8299 
8363 

8241 
8306 
8370 

8248 
8312 
8376 

8254 
8319 
8382 

1  1 
1  1 
1  1 

334 
334 
334 

556 
556 

456 

69 
70 
71 

8388 
8451 
8513 

8395 
8457 
8519 

8401 
8463 
8525 

8407 
8470 
8531 

8414 
8476 
8537 

8420 

8482 
8543 

8426 

8488 
8549 

8432 
8494 
8555 

8439 
8500 
8561 

8445 
8506 
8567 

1  1 
1  1 
1  1 

2  3 
2  3 
2  3 

456 
456 
455 

72 
73 
74 

8573 
8633 
8692 

8579 
8639 
8698 

8585 
8645 
8704 

8591 
8651 
8710 

8597 
8657 
8716 

8603 
8663 

8722 

8609 
8669 

8727 

8615 
8675 
8733 

8621 
8681 
8739 

8627 
8686 
8745 

1  1 
1  1 
1  1 

2  3 
2  3 
234 

455 
455 
455 

75 

8751 

8756 

8762 

8768 

8774 

8779 

8785 

8791 

8797 

8802 

1  1 

233 

455 

76 

77 
78 

8808 
8865 
8921 

8814 
8871 
8927 

8820 
8876 
8932 

8825 
8882 
8938 

8831 
8887 
8943 

8837 
8893 
8949 

8842 
8899 
8954 

8848 
8904 
8960 

8854 
8010 
8965 

8859 
8915 
8971 

1  1 

1 
1 

233 
233 
233 

455 
445 
445 

79 

80 
81 

8976 
9031 
9085 

8982 
9036 
9090 

8987 
9042 
9096 

8993 
9047 
9101 

8998 
9053 
9106 

9004 
9058 
9112 

9009 
9003 
9117 

9015 
9069 
9122 

9020 
9074 
9128 

9025 
9079 
9133 

1 
1 
1 

233 
233 

233 

445 
445 
445 

82 
83 
84 

9138 
9191 
9243 

9143 
9196 
9248 

9149 
9201 
9253 

9154 
9206 
9258 

9159 
9212 
9263 

9165 
9217 
9269 

9170 
9222 

9274 

9175 
9227 
9279 

9180 
9232 
9284 

9186 
9238 
9289 

1 
1 
1  1 

233 
233 
233 

445 
445 
445 

85 

9294 

9299 

9304 

9309 

9315 

9320 

9325 

9330 

9335 

9340 

1  1 

233 

445 

86 

87 
88 

9345 
9395 
9445 

9350 
9400 
9450 

9355 
9405 
9455 

9360 
9410 
9460 

9365 
9415 
9465 

9370 
9420 
9469 

9375 
9425 
9474 

9380 
9430 
9479 

9385 
9435 
9484 

9390 
9440 
9489 

112 
0  1  1 
0  1  1 

233 

223 
223 

445 
344 
344 

89 
90 
91 

9494 
9542 
9590 

9499 
9547 
9595 

9504 
9552 
9600 

9509 
9557 
9605 

9513 
9562 
9609 

9518 
9566 
9614 

9523 
9571 
9619 

9528 
9576 
9624 

9533 
9581 
9628 

9538 
9586 
9633 

0   1 
0   1 
0   1 

223 
223 
223 

344 
344 
344 

92 
93 
94 

9638 
9685 
9731 

9643 
9639 
9736 

9647 
9694 
9741 

9652 
9699 
9745 

9657 
9703 
9750 

9661 
9708 
9754 

9666 
9713 
9759 

9671 
9717 
9763 

9675 
9722 
9768 

9680 
9727 
9773 

0   1 

0   1 
0   1 

223 
223 
223 

344 
344 
344 

95 

9777 

9782 

9786 

9791 

9795 

9800 

9805 

9809 

9814 

9818 

0  1  1 

223 

344 

96 
97 

98 

9823 
9868 
9912 

9827 
9872 
9917 

9832 
9877 
9921 

9836 
9881 
9926 

9841 
9886 
9930 

9845 
9890 
9934 

9850 
9894 
9939 

9854 
9899 
9943 

9859 
9903 
9948 

9863 
9908 
9952 

0  1  1 
0  1  1 
0  1  1 

223 
223 
223 

344 
344 
344 

99 

9956 

9961 

9965 

9969 

9974 

9978 

9983 

9987 

9991 

9996 

0  1  1 

223 

334 

INDEX 


Absolute  pressure,  46 

Absolute  temperature,  12,  14 

Absolute  zero,  13 

Adiabatic,  for  gases,  16, 19,  20,  32  ;  for 

steam,   71 ;    for  superheated   steam, 

97 

Air  compressor,  53 
Air  flow,  113 
Air  refrigeration,  57 
Availability  of  energy,  2,  121 

Battelli,  93 

Boiler  explosion,  77 

Boiler  horse  power,  78 

Boulvin  diagram,  87 

Boyle's  law,  8 

British  thermal  unit,  5 

Calorie,  5 

Calorimeter,  throttling,  119,  121 

Carnot  cycle,  22 ;  for  perfect  gas,  30 
plot  of,  34,  35;  efficiency,  24,  31 
engine,  23  ;  theorem,  24,  26 

Cazin,  71 

Centigrade  scale,  4,  12 

Chappuis,  12 

Charles's  law,  11 

Clausius,  27  ;  inequality  of,  37 

Clearance,  in  air  compressor,  53 ;  ir 
gas  engine,  54 

Compression,  steam  at,  88 

Condenser,  jet,  80  ;  surface,  79 

Conservation  of  energy,  2,  5 

Continuity,  equation  of,  105 

Cycle,  see  Carnot,  reversible,  etc. 

Cylinder  condensation,  90,  98 

Dalton's  law,  44 

De  Laval  turbine,  122 

De  Rochas  gag-engine  cycle,  54 

De  Saint  Venant,  formula  of,  104 

Diesel  internal-combustion  cycle,  56 

Dry  steam,  62 


Efficiency,  see  Carnot,  Rankine,  etc. 
Energy,  1 ;  see  intrinsic,  kinetic,  etc. 
Entropy,  31 ;  measurement  of,  33  ;  for 
41 ;  for  water,  67 ;  for  steam, 
for  superheated  steam,  94 
Equations,  re'sume'  of,  for  gases,  44 ; 
for  steam,  74  ;  for  superheated  steam, 
99  ;  for  flow  of  fluids,  114 
Equivalent  evaporation,  78 

Fahrenheit  scale,  4 
First  law  of  thermodynamics,  5 
Fliegner's  formulas,  113 
Flow  of  fluids,  Chap.  V 
Friction  in  flow  of  fluids,  109 

Gases,  molecular  theory  of,  3  ;  laws  ofr 

8~11 ;  general  equation,  14  ;  perfect, 

10,  38,  40;  imperfect,  44;  see  also 

intrinsic  energy,  entropy,  specific 

;  heat,  etc. 

;      Gay-Lussac,  law  of,  11 


Heat  contents,  see  heat  equations 
Heat    equations,    for    gases,    40 ; 


for 


steam,     65,     67 ;     for     superheated 

steam,  94 
Helium,  13 
Him,  71 
Hydrogen,  13 
Hyperbolic  expansion  line,  91 

Imperfect  gas,  44 

Incomplete  steam  expansion,  81 

Indicator,  18  ;  card,  88 

Injector,  118 

Intrinsic  energy,  2,  16  ;  for  gases,  43  ; 

for  steam,  67  ;  for  superheated  steam, 

96 

Irreversible  process,  35,  37,  109 
Isentropic,  16,  32,  33  ;   for  gases,  41 ; 

for  steam,  71 ;  for  superheated  steam, 

97 


135 


136 


INDEX 


Isoenergic,  16  ;  see  also  intrinsic  energy 
Isothermal,  16,  19,  20 

Joule,  3,  5  ;  law  of,  9  ;  experiment  of, 
10 

Kelvin,  28,  121 

Kinetic  energy,  2  ;  see  molecular  kinetic 

theory 
Kinetic  energy  due  to  expansion,  104, 

109,  114,  115 

Knoblausch  and  Jakob,  93 
Knoblausch,  Linde,  and  Klebe,  93 

Latent  heat,  7 ;  internal,  7 ;  external, 

7  ;  of  vaporization,  61 
Lewicki,  99 
Loss  of  kinetic  energy  due  to  friction, 

109,  116,  1-22 

Mechanical  equivalent  of  heat,  5 
Molecular  kinetic  theory,  3,  8,  11,  13, 

14,  61,  65,  92 
Mollier  diagram,  122 
Motivity,  121 

n,  for  gases,  39,  43  ;  for  steam,  107 
Napier's  formulas,  112 
Newton,  1,  21 
Nozzle  area,  122 

Otto  gas-engine  cycle,  54 

Peabody,  119 

Perfect  gas,  see  gases 

Poly  tropic  transformation,  39,  47 

Porous  plug  experiment,  35 

Potential  energy,  2 

Pressure,  denned,  15  ;  absolute,  46 

Pressure,  coefficient,  11 

Pressure-entropy  plot,  96,  120 

Pressure-volume  plot,  16,  17 

Problems,  see  Contents 

Quality,  66,  96,  119 


Regenerative  cycle,  51 
Regnault,  12,  14,  73 
Reversible  cycle,  24 

Second  law  of  thermodynamics,  21,  26 

Solutions  .of  problems,  see  Contents 

Specific,  use  of  the  term,  14 

Specific  heat,  5  ;  of  gases,  15,  42 ;  of 
water,  63  ;  of  steam,  73 ;  of  super- 
heated steam,  93 

Specific  volume,  of  water,  65  ;  of  steam, 
65,  67 ;  of  superheated  steam,  94 

Steam,  see  under  entropy,  quality,  etc. 

Sterling  hot-air  engine,  51 

Stodola,  99 

Superheated  steam,  93  ;  see  under  en- 
tropy, kinetic  energy,  etc. 

Superheating  in  practice,  98,  102 

Temperature,  3,  12,  14,  28 

Temperature-entropy  plot,  33 ;  for 
steam,  68 ;  for  superheated  steam, 
94  ;  use  of,  86,  122 

Thermal  units,  4 

Thermodynamic  scale  of  temperature, 
28 

Thermometer,  mercury,  3  ;  hydrogen, 
12,  14 

Throttle  control,  121 

Throttling  calorimeter,  119,  121 

Transformation,  see  adiabatic,  isother- 
mal, etc. 

Tumlirz,  93 

Turbine  nozzle,  112,  122 

Vapor,  see  steam 
Vaporization,  61 
Volume,  see  specific  volume 
Volume  coefficient,  11 

Water,     specific     heat,     63 ;     specific 

volume,  65 

Water  vapor,  see  steam 
Wet  steam,  62 
Work  1 ;  calculations  for,  17 


Rankine,  65,  112  ;   Rankine  cycle  and      Zeuner,  formulas  of,  65,'  107,  108 
efficiency,  80 


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